{"id":7324,"date":"2021-06-08T21:55:43","date_gmt":"2021-06-08T21:55:43","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/types-of-chemical-reactions-single-and-double-displacement-reactions\/"},"modified":"2021-09-23T20:00:09","modified_gmt":"2021-09-23T20:00:09","slug":"types-of-chemical-reactions-single-and-double-displacement-reactions","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/types-of-chemical-reactions-single-and-double-displacement-reactions\/","title":{"raw":"Types of Chemical Reactions: Single- and Double-Displacement Reactions","rendered":"Types of Chemical Reactions: Single- and Double-Displacement Reactions"},"content":{"raw":"<div class=\"textbox textbox--learning-objectives\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Learning Objectives<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>Recognize chemical reactions as single-replacement reactions and double-replacement reactions.<\/li>\r\n \t<li>Use the periodic table, an activity series, or solubility rules to predict whether single-replacement reactions or double-replacement reactions will occur.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nUp to now, we have presented chemical reactions as a topic, but we have not discussed how the products of a chemical reaction can be predicted. Here we will begin our study of certain types of chemical reactions that allow us to predict what the products of the reaction will be.\r\n\r\nA [pb_glossary id=\"8031\"]single-replacement reaction[\/pb_glossary]\u00a0is a chemical reaction in which one element is substituted for another element in a compound, generating a new element and a new compound as products. For example:\r\n<p style=\"text-align: center;\">2HCl(aq) +\u00a0Zn(s) \u2192\u00a0ZnCl<sub>2<\/sub>(aq) +\u00a0H<sub>2<\/sub>(g)<\/p>\r\nis an example of a single-replacement reaction. The hydrogen atoms in HCl are replaced by Zn atoms, and in the process a new element\u2014hydrogen\u2014is formed. Another example of a single-replacement reaction is:\r\n<p style=\"text-align: center;\">2NaCl(aq) + F<sub>2<\/sub>(g) \u2192\u00a02NaF(s) +\u00a0Cl<sub>2<\/sub>(g)<\/p>\r\nHere the negatively charged ion changes from chloride to fluoride. A typical characteristic of a single-replacement reaction is that there is one element as a reactant and another element as a product.\r\n\r\nNot all proposed single-replacement reactions will occur between two given reactants. This is most easily demonstrated with fluorine, chlorine, bromine, and iodine. Collectively, these elements are called the <em>halogens<\/em> and are in the next-to-last column on the periodic table (see Figure 4.1 \"Halogens on the Periodic Table\"). The elements on top of the column will replace the elements below them on the periodic table but not the other way around. Thus, the reaction represented by:\r\n<p style=\"text-align: center;\">CaI<sub>2<\/sub>(s) +\u00a0Cl<sub>2<\/sub>(g) \u2192\u00a0CaCl<sub>2<\/sub>(s) +\u00a0I<sub>2<\/sub>(s)<\/p>\r\nThis reaction will occur, but the reaction\r\n<p style=\"text-align: center;\">CaF<sub>2<\/sub>(s) +\u00a0Br<sub>2<\/sub>(\u2113) \u2192\u00a0CaBr<sub>2<\/sub>(s) +\u00a0F<sub>2<\/sub>(g)<\/p>\r\nwill not because bromine is below fluorine on the periodic table. This is just one of many ways the periodic table helps us understand chemistry.\r\n\r\n[caption id=\"attachment_7322\" align=\"aligncenter\" width=\"600\"]<img class=\"wp-image-7322 size-full\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/Halogens-1.png\" alt=\"Hydrogen, fluorine, chlorine, bromine, iodine, and astatine are halogens.\" width=\"600\" height=\"357\" \/> Figure 4.1 \"Halogens on the Periodic Table.\" The halogens are the elements in the next-to-last column on the periodic table.[\/caption]\r\n\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 4.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<h1>Problems<\/h1>\r\nWill a single-replacement reaction occur? If so, identify the products.\r\n<ol>\r\n \t<li>MgCl<sub>2<\/sub> + I<sub>2<\/sub> \u2192 ?<\/li>\r\n \t<li>CaBr<sub>2<\/sub> + F<sub>2<\/sub> \u2192 ?<\/li>\r\n<\/ol>\r\n<h2>Solutions<\/h2>\r\n<ol>\r\n \t<li>Because iodine is below chlorine on the periodic table, a single-replacement reaction will not occur.<\/li>\r\n \t<li>Because fluorine is above bromine on the periodic table, a single-replacement reaction will occur, and the products of the reaction will be CaF<sub>2<\/sub> and Br<sub>2<\/sub>.<\/li>\r\n<\/ol>\r\n<h1>Test Yourself<\/h1>\r\nWill a single-replacement reaction occur? If so, identify the products.\r\n<p style=\"text-align: center;\">FeI<sub>2<\/sub> + Cl<sub>2<\/sub> \u2192 ?<\/p>\r\n\r\n<h2>Answer<\/h2>\r\nYes; FeCl<sub>2<\/sub> and I<sub>2<\/sub>\r\n\r\n<\/div>\r\n<\/div>\r\nChemical reactivity trends are easy to predict when replacing anions in simple ionic compounds\u2014simply use their relative positions on the periodic table. However, when replacing the cations, the trends are not as straightforward. This is partly because there are so many elements that can form cations; an element in one column on the periodic table may replace another element nearby, or it may not. A list called the [pb_glossary id=\"8044\"]activity series[\/pb_glossary]\u00a0does the same thing the periodic table does for halogens: it lists the elements that will replace elements below them in single-replacement reactions. A simple activity series is shown below.\r\n<div class=\"textbox shaded\">\r\n<h1>Activity Series for Cation Replacement in Single-Replacement Reactions<\/h1>\r\n<ul>\r\n \t<li>Li<\/li>\r\n \t<li>K<\/li>\r\n \t<li>Ba<\/li>\r\n \t<li>Sr<\/li>\r\n \t<li>Ca<\/li>\r\n \t<li>Na<\/li>\r\n \t<li>Mg<\/li>\r\n \t<li>Al<\/li>\r\n \t<li>Mn<\/li>\r\n \t<li>Zn<\/li>\r\n \t<li>Cr<\/li>\r\n \t<li>Fe<\/li>\r\n \t<li>Ni<\/li>\r\n \t<li>Sn<\/li>\r\n \t<li>Pb<\/li>\r\n \t<li>H<sub class=\"subscript\">2<\/sub><\/li>\r\n \t<li>Cu<\/li>\r\n \t<li>Hg<\/li>\r\n \t<li>Ag<\/li>\r\n \t<li>Pd<\/li>\r\n \t<li>Pt<\/li>\r\n \t<li>Au<\/li>\r\n<\/ul>\r\n<\/div>\r\nUsing the activity series is similar to using the positions of the halogens on the periodic table. An element on top will replace an element below it in compounds undergoing a single-replacement reaction. Elements will not replace elements above them in compounds.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 4.3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<h1>Problems<\/h1>\r\nUse the activity series to predict the products, if any, of each equation.\r\n<ol>\r\n \t<li>FeCl<sub>2<\/sub> + Zn \u2192 ?<\/li>\r\n \t<li>HNO<sub>3<\/sub> + Au \u2192 ?<\/li>\r\n<\/ol>\r\n<h2>Solutions<\/h2>\r\n<ol>\r\n \t<li>Because zinc is above iron in the activity series, it will replace iron in the compound. The products of this single-replacement reaction are ZnCl<sub>2<\/sub> and Fe.<\/li>\r\n \t<li>Gold is below hydrogen in the activity series. As such, it will not replace hydrogen in a compound with the nitrate ion. No reaction is predicted.<\/li>\r\n<\/ol>\r\n<h1>Test Yourself<\/h1>\r\nUse the activity series to predict the products, if any, of this equation.\r\n<p style=\"text-align: center;\">AlPO<sub>4<\/sub> + Mg \u2192 ?<\/p>\r\n\r\n<h2>Answer<\/h2>\r\nMg<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> and Al\r\n\r\n<\/div>\r\n<\/div>\r\nA [pb_glossary id=\"8048\"]double-replacement reaction[\/pb_glossary]\u00a0occurs when parts of two ionic compounds are exchanged, making two new compounds. A characteristic of a double-replacement equation is that there are two compounds as reactants and two different compounds as products. An example is:\r\n<p style=\"text-align: center;\">CuCl<sub>2<\/sub>(aq) +\u00a02AgNO<sub>3<\/sub>(aq) \u2192\u00a0Cu(NO<sub>3<\/sub>)<sub>2<\/sub>(aq) +\u00a02AgCl(s)<\/p>\r\nThere are two equivalent ways of considering a double-replacement equation: either the cations are swapped, or the anions are swapped. (You cannot swap both; you would end up with the same substances you started with.) Either perspective should allow you to predict the proper products, as long as you pair a cation with an anion and not a cation with a cation or an anion with an anion.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 4.4<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<h1>Problem<\/h1>\r\nPredict the products of this double-replacement equation: BaCl<sub>2<\/sub> + Na<sub>2<\/sub>SO<sub>4<\/sub> \u2192 ?\r\n<h2>Solution<\/h2>\r\nThinking about the reaction as either switching the cations or switching the anions, we would expect the products to be BaSO<sub>4<\/sub> and NaCl.\r\n<h1>Test Yourself<\/h1>\r\nPredict the products of this double-replacement equation: KBr + AgNO<sub>3<\/sub> \u2192 ?\r\n<h2>Answer<\/h2>\r\nKNO<sub>3<\/sub> and AgBr\r\n\r\n<\/div>\r\n<\/div>\r\nPredicting whether a double-replacement reaction occurs is somewhat more difficult than predicting a single-replacement reaction. However, there is one type of double-replacement reaction that we can predict: the precipitation reaction. A [pb_glossary id=\"8046\"]precipitation reaction[\/pb_glossary]\u00a0occurs when two ionic compounds are dissolved in water and form a new ionic compound that does not dissolve; this new compound falls out of solution as a solid [pb_glossary id=\"8047\"]precipitate[\/pb_glossary]. The formation of a solid precipitate is the driving force that makes the reaction proceed.\r\n\r\nTo judge whether double-replacement reactions will occur, we need to know what kinds of ionic compounds form precipitates. For this, we use [pb_glossary id=\"8049\"]solubility rules[\/pb_glossary], which are general statements that predict which ionic compounds dissolve (are soluble) and which do not (are not soluble or insoluble). Tables 4.1a and b \"Some Useful Solubility Rules\" list some general solubility rules. We need to consider each ionic compound (both the reactants and the possible products) in light of the solubility rules in Table 4.1a and b. If a compound is soluble, we use the (aq) label with it, indicating it dissolves. If a compound is not soluble, we use the (s) label with it and assume that it will precipitate out of solution. If everything is soluble, then no reaction will be expected.\r\n<table id=\"tab4.1a\" class=\"aligncenter\" style=\"border-spacing: 0px;\" cellspacing=\"0px\" cellpadding=\"0\"><caption>Table 4.1a Some Useful Solubility Rules<\/caption>\r\n<tbody>\r\n<tr>\r\n<th scope=\"col\">These compounds generally dissolve in water (are soluble):<\/th>\r\n<th scope=\"col\">Exceptions:<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>All compounds of Li<sup>+<\/sup>, Na<sup>+<\/sup>, K<sup>+<\/sup>, Rb<sup>+<\/sup>, Cs<sup>+<\/sup>, and NH<sub>4<\/sub><sup>+<\/sup><\/td>\r\n<td>None<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>All compounds of NO<sub>3<\/sub><sup>\u2212<\/sup> and C<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub><sup>\u2212<\/sup><\/td>\r\n<td>None<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Compounds of Cl<sup>\u2212<\/sup>, Br<sup>\u2212<\/sup>, I<sup>\u2212<\/sup><\/td>\r\n<td>Ag<sup>+<\/sup>, Hg<sub>2<\/sub><sup>2+<\/sup>, Pb<sup>2+<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Compounds of SO<sub>4<\/sub><sup>2<\/sup><\/td>\r\n<td>Hg<sub>2<\/sub><sup>2+<\/sup>, Pb<sup>2+<\/sup>, Sr<sup>2+<\/sup>, Ba<sup>2+<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<table id=\"tab4.1b\" class=\"aligncenter\" style=\"border-collapse: collapse; width: 99.8647%;\" border=\"0\"><caption>Table 4.1b Some More Useful Solubility Rules<\/caption>\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 49.7921%;\" scope=\"col\">These compounds generally do not dissolve in water (are insoluble):<\/th>\r\n<th style=\"width: 50.0727%;\" scope=\"col\">Exceptions:<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 49.7921%;\">Compounds of CO<sub>3<\/sub><sup>2\u2212<\/sup> and PO<sub>4<\/sub><sup>3\u2212<\/sup><\/td>\r\n<td style=\"width: 50.0727%;\">Compounds of Li<sup>+<\/sup>, Na<sup>+<\/sup>, K<sup>+<\/sup>, Rb<sup>+<\/sup>, Cs<sup>+<\/sup>, and NH<sub>4<\/sub><sup>+<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 49.7921%;\">Compounds of OH<sup>\u2212<\/sup><\/td>\r\n<td style=\"width: 50.0727%;\">Compounds of Li<sup>+<\/sup>, Na<sup>+<\/sup>, K<sup>+<\/sup>, Rb<sup>+<\/sup>, Cs<sup>+<\/sup>, NH<sub>4<\/sub><sup>+<\/sup>, Sr<sup>2+<\/sup>, and Ba<sup>2+<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFor example, consider the possible double-replacement reaction between Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> and SrCl<sub class=\"subscript\">2<\/sub>. The solubility rules say that all ionic sodium compounds are soluble and all ionic chloride compounds are soluble except for Ag<sup class=\"superscript\">+<\/sup>, Hg<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">2+<\/sup>, and Pb<sup class=\"superscript\">2+<\/sup>, which are not being considered here. Therefore, Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> and SrCl<sub class=\"subscript\">2<\/sub> are both soluble. The possible double-replacement reaction products are NaCl and SrSO<sub class=\"subscript\">4<\/sub>. Are these soluble? NaCl is (by the same rule we just quoted), but what about SrSO<sub class=\"subscript\">4<\/sub>? Compounds of the sulfate ion are generally soluble, but Sr<sup class=\"superscript\">2+<\/sup> is an exception: we expect it to be insoluble\u2014a precipitate. Therefore, we expect a reaction to occur, and the balanced chemical equation would be:\r\n<p style=\"text-align: center;\">Na<sub>2<\/sub>SO<sub>4<\/sub>(aq) +\u00a0SrCl<sub>2<\/sub>(aq) \u2192 2NaCl(aq) + SrSO<sub>4<\/sub>(s)<\/p>\r\nYou would expect to see a visual change corresponding to SrSO<sub>4<\/sub> precipitating out of solution (Figure 4.2 \"Double-Replacement Reactions\").\r\n\r\n[caption id=\"attachment_7323\" align=\"aligncenter\" width=\"350\"]<img class=\"wp-image-7323\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/Double-Replacement-Reaction-1.png\" alt=\"A beaker full of a blue liquid. Some white solids are at the bottom.\" width=\"350\" height=\"373\" \/> Figure 4.2 \"Double-Replacement Reactions.\" Some double-replacement reactions are obvious because you can see a solid precipitate coming out of the solution.[\/caption]\r\n\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 4.5<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<h1>Problems<\/h1>\r\nWill a double-replacement reaction occur? If so, identify the products.\r\n<ol>\r\n \t<li>Ca(NO<sub>3<\/sub>)<sub>2<\/sub> + KBr \u2192 ?<\/li>\r\n \t<li>NaOH + FeCl<sub>2<\/sub> \u2192 ?<\/li>\r\n<\/ol>\r\n<h2>Solutions<\/h2>\r\n<ol>\r\n \t<li>According to the solubility rules, both Ca(NO<sub>3<\/sub>)<sub>2<\/sub> and KBr are soluble. Now we consider what the double-replacement products would be by switching the cations (or the anions)\u2014namely, CaBr<sub>2<\/sub> and KNO<sub>3<\/sub>. However, the solubility rules predict that these two substances would also be soluble, so no precipitate would form. Thus, we predict no reaction in this case.<\/li>\r\n \t<li>According to the solubility rules, both NaOH and FeCl<sub>2<\/sub> are expected to be soluble. If we assume that a double-replacement reaction may occur, we need to consider the possible products, which would be NaCl and Fe(OH)<sub>2<\/sub>. NaCl is soluble, but, according to the solubility rules, Fe(OH)<sub>2<\/sub> is not. Therefore, a reaction would occur, and Fe(OH)<sub>2<\/sub>(s) would precipitate out of solution. The balanced chemical equation is: 2NaOH(aq) + FeCl<sub>2<\/sub>(aq) \u2192 2NaCl(aq) + Fe(OH)<sub>2<\/sub>(s)<\/li>\r\n<\/ol>\r\n<h1>Test Yourself<\/h1>\r\nWill a double-replacement equation occur? If so, identify the products.\r\n<p style=\"text-align: center;\">Sr(NO<sub>3<\/sub>)<sub>2<\/sub> + KCl \u2192 ?<\/p>\r\n\r\n<h2>Answer<\/h2>\r\nNo reaction; all possible products are soluble.\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Key Takeaways<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ul>\r\n \t<li>A single-replacement reaction replaces one element for another in a compound.<\/li>\r\n \t<li>The periodic table or an activity series can help predict whether single-replacement reactions occur.<\/li>\r\n \t<li>A double-replacement reaction exchanges the cations (or the anions) of two ionic compounds.<\/li>\r\n \t<li>A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate.<\/li>\r\n \t<li>Solubility rules are used to predict whether some double-replacement reactions will occur.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercises<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<h1>Questions<\/h1>\r\n<ol>\r\n \t<li>What are the general characteristics that help you recognize single-replacement reactions?<\/li>\r\n \t<li>What are the general characteristics that help you recognize double-replacement reactions?<\/li>\r\n \t<li>Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.\r\n<ol type=\"a\">\r\n \t<li>Zn + Fe(NO<sub>3<\/sub>)<sub>2<\/sub> \u2192 ?<\/li>\r\n \t<li>F<sub>2<\/sub> + FeI<sub>3<\/sub> \u2192 ?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.\r\n<ol type=\"a\">\r\n \t<li>Li + MgSO<sub>4<\/sub> \u2192 ?<\/li>\r\n \t<li>NaBr + Cl<sub>2<\/sub> \u2192 ?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.\r\n<ol type=\"a\">\r\n \t<li>Sn + H<sub>2<\/sub>SO<sub>4<\/sub> \u2192 ?<\/li>\r\n \t<li>Al + NiBr<sub>2<\/sub> \u2192 ?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.\r\n<ol type=\"a\">\r\n \t<li>Mg + HCl \u2192 ?<\/li>\r\n \t<li>HI + Br<sub>2<\/sub> \u2192 ?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.\r\n<ol type=\"a\">\r\n \t<li>FeCl<sub>2<\/sub> + Br<sub>2<\/sub> \u2192 ?<\/li>\r\n \t<li>Fe(NO<sub>3<\/sub>)<sub>3<\/sub> + Al \u2192 ?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.\r\n<ol type=\"a\">\r\n \t<li>Zn + Fe<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> \u2192 ?<\/li>\r\n \t<li>Ag + HNO<sub>3<\/sub> \u2192 ?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.\r\n<ol type=\"a\">\r\n \t<li>NaI + Cl<sub>2<\/sub> \u2192 ?<\/li>\r\n \t<li>AgCl + Au \u2192 ?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.\r\n<ol type=\"a\">\r\n \t<li>Pt + H<sub>3<\/sub>PO<sub>4<\/sub> \u2192 ?<\/li>\r\n \t<li>Li + H<sub>2<\/sub>O \u2192 ? (Hint: treat H<sub>2<\/sub>O as if it were composed of H<sup>+<\/sup> and OH<sup>\u2212<\/sup> ions.)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.\r\n<ol type=\"a\">\r\n \t<li>Zn(NO<sub>3<\/sub>)<sub>2<\/sub> + NaOH \u2192 ?<\/li>\r\n \t<li>HCl + Na<sub>2<\/sub>S \u2192 ?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.\r\n<ol type=\"a\">\r\n \t<li>Ca(C<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub>)<sub>2<\/sub> + HNO<sub>3<\/sub> \u2192 ?<\/li>\r\n \t<li>Na<sub>2<\/sub>CO<sub>3<\/sub> + Sr(NO<sub>2<\/sub>)<sub>2<\/sub> \u2192 ?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.\r\n<ol type=\"a\">\r\n \t<li>Pb(NO<sub>3<\/sub>)<sub>2<\/sub> + KBr \u2192 ?<\/li>\r\n \t<li>K<sub>2<\/sub>O + MgCO<sub>3<\/sub> \u2192 ?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.\r\n<ol type=\"a\">\r\n \t<li>Sn(OH)<sub>2<\/sub> + FeBr<sub>3<\/sub> \u2192 ?<\/li>\r\n \t<li>CsNO<sub>3<\/sub> + KCl \u2192 ?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.\r\n<ol type=\"a\">\r\n \t<li>Pb(NO<sub>3<\/sub>)<sub>2<\/sub> + KBr \u2192 ?<\/li>\r\n \t<li>K<sub>2<\/sub>O + Na<sub>2<\/sub>CO<sub>3<\/sub> \u2192 ?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.\r\n<ol type=\"a\">\r\n \t<li>Na<sub>2<\/sub>CO<sub>3<\/sub> + Sr(NO<sub>2<\/sub>)<sub>2<\/sub> \u2192 ?<\/li>\r\n \t<li>(NH<sub>4<\/sub>)<sub>2<\/sub>SO<sub>4<\/sub> + Ba(NO<sub>3<\/sub>)<sub>2<\/sub> \u2192 ?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.\r\n<ol type=\"a\">\r\n \t<li>K<sub>3<\/sub>PO<sub>4<\/sub> + SrCl<sub>2<\/sub> \u2192 ?<\/li>\r\n \t<li>NaOH + MgCl<sub>2<\/sub> \u2192 ?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.\r\n<ol type=\"a\">\r\n \t<li>KC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub> + Li<sub>2<\/sub>CO<sub>3<\/sub> \u2192 ?<\/li>\r\n \t<li>KOH + AgNO<sub>3<\/sub> \u2192 ?<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<h1>Answers<\/h1>\r\n<ol>\r\n \t<li>One element replaces another element in a compound.<\/li>\r\n<\/ol>\r\n<ol start=\"3\">\r\n \t<li>\r\n<ol type=\"a\">\r\n \t<li>Zn + Fe(NO<sub>3<\/sub>)<sub>2<\/sub> \u2192 Zn(NO<sub>3<\/sub>)<sub>2<\/sub> + Fe<\/li>\r\n \t<li>3F<sub>2<\/sub> + 2FeI<sub>3<\/sub> \u2192 3I<sub>2<\/sub> + 2FeF<sub>3<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<ol start=\"5\">\r\n \t<li>\r\n<ol type=\"a\">\r\n \t<li>Sn + H<sub>2<\/sub>SO<sub>4<\/sub> \u2192 SnSO<sub>4<\/sub> + H<sub>2<\/sub><\/li>\r\n \t<li>2Al + 3NiBr<sub>2<\/sub> \u2192 2AlBr<sub>3<\/sub> + 3Ni<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<ol start=\"7\">\r\n \t<li>\r\n<ol type=\"a\">\r\n \t<li>No reaction occurs.<\/li>\r\n \t<li>Fe(NO<sub>3<\/sub>)<sub>3<\/sub> + Al \u2192 Al(NO<sub>3<\/sub>)<sub>3<\/sub> + Fe<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<ol start=\"9\">\r\n \t<li>\r\n<ol type=\"a\">\r\n \t<li>2NaI + Cl<sub>2<\/sub> \u2192 2NaCl + I<sub>2<\/sub><\/li>\r\n \t<li>No reaction occurs.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<ol start=\"11\">\r\n \t<li>\r\n<ol type=\"a\">\r\n \t<li>Zn(NO<sub>3<\/sub>)<sub>2 <\/sub>+ 2NaOH \u2192 Zn(OH)<sub>2 <\/sub>+ 2NaNO<sub>3<\/sub><\/li>\r\n \t<li>2HCl + Na<sub>2<\/sub>S \u2192 2NaCl + H<sub>2<\/sub>S<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<ol start=\"13\">\r\n \t<li>\r\n<ol type=\"a\">\r\n \t<li>Pb(NO<sub>3<\/sub>)<sub>2<\/sub> + 2KBr \u2192 PbBr<sub>2<\/sub> + 2KNO<sub>3<\/sub><\/li>\r\n \t<li>K<sub>2<\/sub>O + MgCO<sub>3<\/sub> \u2192 K<sub>2<\/sub>CO<sub>3<\/sub> + MgO<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<ol start=\"15\">\r\n \t<li>\r\n<ol type=\"a\">\r\n \t<li>Pb(NO<sub>3<\/sub>)<sub>2<\/sub> + 2KBr \u2192 PbBr<sub>2<\/sub>(s) + 2KNO<sub>3<\/sub><\/li>\r\n \t<li>No reaction occurs.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<ol start=\"17\">\r\n \t<li>\r\n<ol type=\"a\">\r\n \t<li>2K<sub>3<\/sub>PO<sub>4<\/sub> + 3SrCl<sub>2<\/sub> \u2192 Sr<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>(s) + 6KCl<\/li>\r\n \t<li>2NaOH + MgCl<sub>2<\/sub> \u2192 2NaCl + Mg(OH)<sub>2<\/sub>(s)<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div id=\"ball-ch04_s02_n07\" class=\"key_takeaways editable block\">\r\n<h3>Media Attributions<\/h3>\r\nFigure 4.1\r\n<ul>\r\n \t<li>\"Halogens on the Periodic Table\" by David W. Ball \u00a9 <a href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/3.0\/\">CC BY-NC-SA (Attribution NonCommercial ShareAlike)<\/a><\/li>\r\n<\/ul>\r\nFigure 4.2\r\n<ul>\r\n \t<li><a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Copper_solution.jpg\">\"Copper solution\"<\/a> by Choij \u00a9 <a href=\"https:\/\/creativecommons.org\/publicdomain\/mark\/1.0\/\">Public Domain<\/a><\/li>\r\n<\/ul>\r\n<\/div>","rendered":"<div class=\"textbox textbox--learning-objectives\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Learning Objectives<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>Recognize chemical reactions as single-replacement reactions and double-replacement reactions.<\/li>\n<li>Use the periodic table, an activity series, or solubility rules to predict whether single-replacement reactions or double-replacement reactions will occur.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p>Up to now, we have presented chemical reactions as a topic, but we have not discussed how the products of a chemical reaction can be predicted. Here we will begin our study of certain types of chemical reactions that allow us to predict what the products of the reaction will be.<\/p>\n<p>A <a class=\"glossary-term\" aria-haspopup=\"dialog\" aria-describedby=\"definition\" href=\"#term_7324_8031\">single-replacement reaction<\/a>\u00a0is a chemical reaction in which one element is substituted for another element in a compound, generating a new element and a new compound as products. For example:<\/p>\n<p style=\"text-align: center;\">2HCl(aq) +\u00a0Zn(s) \u2192\u00a0ZnCl<sub>2<\/sub>(aq) +\u00a0H<sub>2<\/sub>(g)<\/p>\n<p>is an example of a single-replacement reaction. The hydrogen atoms in HCl are replaced by Zn atoms, and in the process a new element\u2014hydrogen\u2014is formed. Another example of a single-replacement reaction is:<\/p>\n<p style=\"text-align: center;\">2NaCl(aq) + F<sub>2<\/sub>(g) \u2192\u00a02NaF(s) +\u00a0Cl<sub>2<\/sub>(g)<\/p>\n<p>Here the negatively charged ion changes from chloride to fluoride. A typical characteristic of a single-replacement reaction is that there is one element as a reactant and another element as a product.<\/p>\n<p>Not all proposed single-replacement reactions will occur between two given reactants. This is most easily demonstrated with fluorine, chlorine, bromine, and iodine. Collectively, these elements are called the <em>halogens<\/em> and are in the next-to-last column on the periodic table (see Figure 4.1 &#8220;Halogens on the Periodic Table&#8221;). The elements on top of the column will replace the elements below them on the periodic table but not the other way around. Thus, the reaction represented by:<\/p>\n<p style=\"text-align: center;\">CaI<sub>2<\/sub>(s) +\u00a0Cl<sub>2<\/sub>(g) \u2192\u00a0CaCl<sub>2<\/sub>(s) +\u00a0I<sub>2<\/sub>(s)<\/p>\n<p>This reaction will occur, but the reaction<\/p>\n<p style=\"text-align: center;\">CaF<sub>2<\/sub>(s) +\u00a0Br<sub>2<\/sub>(\u2113) \u2192\u00a0CaBr<sub>2<\/sub>(s) +\u00a0F<sub>2<\/sub>(g)<\/p>\n<p>will not because bromine is below fluorine on the periodic table. This is just one of many ways the periodic table helps us understand chemistry.<\/p>\n<figure id=\"attachment_7322\" aria-describedby=\"caption-attachment-7322\" style=\"width: 600px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-7322 size-full\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/Halogens-1.png\" alt=\"Hydrogen, fluorine, chlorine, bromine, iodine, and astatine are halogens.\" width=\"600\" height=\"357\" srcset=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/Halogens-1.png 600w, https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/Halogens-1-300x179.png 300w, https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/Halogens-1-65x39.png 65w, https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/Halogens-1-225x134.png 225w, https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/Halogens-1-350x208.png 350w\" sizes=\"auto, (max-width: 600px) 100vw, 600px\" \/><figcaption id=\"caption-attachment-7322\" class=\"wp-caption-text\">Figure 4.1 &#8220;Halogens on the Periodic Table.&#8221; The halogens are the elements in the next-to-last column on the periodic table.<\/figcaption><\/figure>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 4.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<h1>Problems<\/h1>\n<p>Will a single-replacement reaction occur? If so, identify the products.<\/p>\n<ol>\n<li>MgCl<sub>2<\/sub> + I<sub>2<\/sub> \u2192 ?<\/li>\n<li>CaBr<sub>2<\/sub> + F<sub>2<\/sub> \u2192 ?<\/li>\n<\/ol>\n<h2>Solutions<\/h2>\n<ol>\n<li>Because iodine is below chlorine on the periodic table, a single-replacement reaction will not occur.<\/li>\n<li>Because fluorine is above bromine on the periodic table, a single-replacement reaction will occur, and the products of the reaction will be CaF<sub>2<\/sub> and Br<sub>2<\/sub>.<\/li>\n<\/ol>\n<h1>Test Yourself<\/h1>\n<p>Will a single-replacement reaction occur? If so, identify the products.<\/p>\n<p style=\"text-align: center;\">FeI<sub>2<\/sub> + Cl<sub>2<\/sub> \u2192 ?<\/p>\n<h2>Answer<\/h2>\n<p>Yes; FeCl<sub>2<\/sub> and I<sub>2<\/sub><\/p>\n<\/div>\n<\/div>\n<p>Chemical reactivity trends are easy to predict when replacing anions in simple ionic compounds\u2014simply use their relative positions on the periodic table. However, when replacing the cations, the trends are not as straightforward. This is partly because there are so many elements that can form cations; an element in one column on the periodic table may replace another element nearby, or it may not. A list called the <a class=\"glossary-term\" aria-haspopup=\"dialog\" aria-describedby=\"definition\" href=\"#term_7324_8044\">activity series<\/a>\u00a0does the same thing the periodic table does for halogens: it lists the elements that will replace elements below them in single-replacement reactions. A simple activity series is shown below.<\/p>\n<div class=\"textbox shaded\">\n<h1>Activity Series for Cation Replacement in Single-Replacement Reactions<\/h1>\n<ul>\n<li>Li<\/li>\n<li>K<\/li>\n<li>Ba<\/li>\n<li>Sr<\/li>\n<li>Ca<\/li>\n<li>Na<\/li>\n<li>Mg<\/li>\n<li>Al<\/li>\n<li>Mn<\/li>\n<li>Zn<\/li>\n<li>Cr<\/li>\n<li>Fe<\/li>\n<li>Ni<\/li>\n<li>Sn<\/li>\n<li>Pb<\/li>\n<li>H<sub class=\"subscript\">2<\/sub><\/li>\n<li>Cu<\/li>\n<li>Hg<\/li>\n<li>Ag<\/li>\n<li>Pd<\/li>\n<li>Pt<\/li>\n<li>Au<\/li>\n<\/ul>\n<\/div>\n<p>Using the activity series is similar to using the positions of the halogens on the periodic table. An element on top will replace an element below it in compounds undergoing a single-replacement reaction. Elements will not replace elements above them in compounds.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 4.3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<h1>Problems<\/h1>\n<p>Use the activity series to predict the products, if any, of each equation.<\/p>\n<ol>\n<li>FeCl<sub>2<\/sub> + Zn \u2192 ?<\/li>\n<li>HNO<sub>3<\/sub> + Au \u2192 ?<\/li>\n<\/ol>\n<h2>Solutions<\/h2>\n<ol>\n<li>Because zinc is above iron in the activity series, it will replace iron in the compound. The products of this single-replacement reaction are ZnCl<sub>2<\/sub> and Fe.<\/li>\n<li>Gold is below hydrogen in the activity series. As such, it will not replace hydrogen in a compound with the nitrate ion. No reaction is predicted.<\/li>\n<\/ol>\n<h1>Test Yourself<\/h1>\n<p>Use the activity series to predict the products, if any, of this equation.<\/p>\n<p style=\"text-align: center;\">AlPO<sub>4<\/sub> + Mg \u2192 ?<\/p>\n<h2>Answer<\/h2>\n<p>Mg<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> and Al<\/p>\n<\/div>\n<\/div>\n<p>A <a class=\"glossary-term\" aria-haspopup=\"dialog\" aria-describedby=\"definition\" href=\"#term_7324_8048\">double-replacement reaction<\/a>\u00a0occurs when parts of two ionic compounds are exchanged, making two new compounds. A characteristic of a double-replacement equation is that there are two compounds as reactants and two different compounds as products. An example is:<\/p>\n<p style=\"text-align: center;\">CuCl<sub>2<\/sub>(aq) +\u00a02AgNO<sub>3<\/sub>(aq) \u2192\u00a0Cu(NO<sub>3<\/sub>)<sub>2<\/sub>(aq) +\u00a02AgCl(s)<\/p>\n<p>There are two equivalent ways of considering a double-replacement equation: either the cations are swapped, or the anions are swapped. (You cannot swap both; you would end up with the same substances you started with.) Either perspective should allow you to predict the proper products, as long as you pair a cation with an anion and not a cation with a cation or an anion with an anion.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 4.4<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<h1>Problem<\/h1>\n<p>Predict the products of this double-replacement equation: BaCl<sub>2<\/sub> + Na<sub>2<\/sub>SO<sub>4<\/sub> \u2192 ?<\/p>\n<h2>Solution<\/h2>\n<p>Thinking about the reaction as either switching the cations or switching the anions, we would expect the products to be BaSO<sub>4<\/sub> and NaCl.<\/p>\n<h1>Test Yourself<\/h1>\n<p>Predict the products of this double-replacement equation: KBr + AgNO<sub>3<\/sub> \u2192 ?<\/p>\n<h2>Answer<\/h2>\n<p>KNO<sub>3<\/sub> and AgBr<\/p>\n<\/div>\n<\/div>\n<p>Predicting whether a double-replacement reaction occurs is somewhat more difficult than predicting a single-replacement reaction. However, there is one type of double-replacement reaction that we can predict: the precipitation reaction. A <a class=\"glossary-term\" aria-haspopup=\"dialog\" aria-describedby=\"definition\" href=\"#term_7324_8046\">precipitation reaction<\/a>\u00a0occurs when two ionic compounds are dissolved in water and form a new ionic compound that does not dissolve; this new compound falls out of solution as a solid <a class=\"glossary-term\" aria-haspopup=\"dialog\" aria-describedby=\"definition\" href=\"#term_7324_8047\">precipitate<\/a>. The formation of a solid precipitate is the driving force that makes the reaction proceed.<\/p>\n<p>To judge whether double-replacement reactions will occur, we need to know what kinds of ionic compounds form precipitates. For this, we use <a class=\"glossary-term\" aria-haspopup=\"dialog\" aria-describedby=\"definition\" href=\"#term_7324_8049\">solubility rules<\/a>, which are general statements that predict which ionic compounds dissolve (are soluble) and which do not (are not soluble or insoluble). Tables 4.1a and b &#8220;Some Useful Solubility Rules&#8221; list some general solubility rules. We need to consider each ionic compound (both the reactants and the possible products) in light of the solubility rules in Table 4.1a and b. If a compound is soluble, we use the (aq) label with it, indicating it dissolves. If a compound is not soluble, we use the (s) label with it and assume that it will precipitate out of solution. If everything is soluble, then no reaction will be expected.<\/p>\n<table id=\"tab4.1a\" class=\"aligncenter\" style=\"border-spacing: 0px; border-spacing: 0pxpx;\" cellpadding=\"0\">\n<caption>Table 4.1a Some Useful Solubility Rules<\/caption>\n<tbody>\n<tr>\n<th scope=\"col\">These compounds generally dissolve in water (are soluble):<\/th>\n<th scope=\"col\">Exceptions:<\/th>\n<\/tr>\n<tr>\n<td>All compounds of Li<sup>+<\/sup>, Na<sup>+<\/sup>, K<sup>+<\/sup>, Rb<sup>+<\/sup>, Cs<sup>+<\/sup>, and NH<sub>4<\/sub><sup>+<\/sup><\/td>\n<td>None<\/td>\n<\/tr>\n<tr>\n<td>All compounds of NO<sub>3<\/sub><sup>\u2212<\/sup> and C<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub><sup>\u2212<\/sup><\/td>\n<td>None<\/td>\n<\/tr>\n<tr>\n<td>Compounds of Cl<sup>\u2212<\/sup>, Br<sup>\u2212<\/sup>, I<sup>\u2212<\/sup><\/td>\n<td>Ag<sup>+<\/sup>, Hg<sub>2<\/sub><sup>2+<\/sup>, Pb<sup>2+<\/sup><\/td>\n<\/tr>\n<tr>\n<td>Compounds of SO<sub>4<\/sub><sup>2<\/sup><\/td>\n<td>Hg<sub>2<\/sub><sup>2+<\/sup>, Pb<sup>2+<\/sup>, Sr<sup>2+<\/sup>, Ba<sup>2+<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table id=\"tab4.1b\" class=\"aligncenter\" style=\"border-collapse: collapse; width: 99.8647%;\">\n<caption>Table 4.1b Some More Useful Solubility Rules<\/caption>\n<tbody>\n<tr>\n<th style=\"width: 49.7921%;\" scope=\"col\">These compounds generally do not dissolve in water (are insoluble):<\/th>\n<th style=\"width: 50.0727%;\" scope=\"col\">Exceptions:<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 49.7921%;\">Compounds of CO<sub>3<\/sub><sup>2\u2212<\/sup> and PO<sub>4<\/sub><sup>3\u2212<\/sup><\/td>\n<td style=\"width: 50.0727%;\">Compounds of Li<sup>+<\/sup>, Na<sup>+<\/sup>, K<sup>+<\/sup>, Rb<sup>+<\/sup>, Cs<sup>+<\/sup>, and NH<sub>4<\/sub><sup>+<\/sup><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 49.7921%;\">Compounds of OH<sup>\u2212<\/sup><\/td>\n<td style=\"width: 50.0727%;\">Compounds of Li<sup>+<\/sup>, Na<sup>+<\/sup>, K<sup>+<\/sup>, Rb<sup>+<\/sup>, Cs<sup>+<\/sup>, NH<sub>4<\/sub><sup>+<\/sup>, Sr<sup>2+<\/sup>, and Ba<sup>2+<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>For example, consider the possible double-replacement reaction between Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> and SrCl<sub class=\"subscript\">2<\/sub>. The solubility rules say that all ionic sodium compounds are soluble and all ionic chloride compounds are soluble except for Ag<sup class=\"superscript\">+<\/sup>, Hg<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">2+<\/sup>, and Pb<sup class=\"superscript\">2+<\/sup>, which are not being considered here. Therefore, Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> and SrCl<sub class=\"subscript\">2<\/sub> are both soluble. The possible double-replacement reaction products are NaCl and SrSO<sub class=\"subscript\">4<\/sub>. Are these soluble? NaCl is (by the same rule we just quoted), but what about SrSO<sub class=\"subscript\">4<\/sub>? Compounds of the sulfate ion are generally soluble, but Sr<sup class=\"superscript\">2+<\/sup> is an exception: we expect it to be insoluble\u2014a precipitate. Therefore, we expect a reaction to occur, and the balanced chemical equation would be:<\/p>\n<p style=\"text-align: center;\">Na<sub>2<\/sub>SO<sub>4<\/sub>(aq) +\u00a0SrCl<sub>2<\/sub>(aq) \u2192 2NaCl(aq) + SrSO<sub>4<\/sub>(s)<\/p>\n<p>You would expect to see a visual change corresponding to SrSO<sub>4<\/sub> precipitating out of solution (Figure 4.2 &#8220;Double-Replacement Reactions&#8221;).<\/p>\n<figure id=\"attachment_7323\" aria-describedby=\"caption-attachment-7323\" style=\"width: 350px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-7323\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/Double-Replacement-Reaction-1.png\" alt=\"A beaker full of a blue liquid. Some white solids are at the bottom.\" width=\"350\" height=\"373\" srcset=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/Double-Replacement-Reaction-1.png 600w, https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/Double-Replacement-Reaction-1-282x300.png 282w, https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/Double-Replacement-Reaction-1-65x69.png 65w, https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/Double-Replacement-Reaction-1-225x240.png 225w, https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/Double-Replacement-Reaction-1-350x373.png 350w\" sizes=\"auto, (max-width: 350px) 100vw, 350px\" \/><figcaption id=\"caption-attachment-7323\" class=\"wp-caption-text\">Figure 4.2 &#8220;Double-Replacement Reactions.&#8221; Some double-replacement reactions are obvious because you can see a solid precipitate coming out of the solution.<\/figcaption><\/figure>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 4.5<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<h1>Problems<\/h1>\n<p>Will a double-replacement reaction occur? If so, identify the products.<\/p>\n<ol>\n<li>Ca(NO<sub>3<\/sub>)<sub>2<\/sub> + KBr \u2192 ?<\/li>\n<li>NaOH + FeCl<sub>2<\/sub> \u2192 ?<\/li>\n<\/ol>\n<h2>Solutions<\/h2>\n<ol>\n<li>According to the solubility rules, both Ca(NO<sub>3<\/sub>)<sub>2<\/sub> and KBr are soluble. Now we consider what the double-replacement products would be by switching the cations (or the anions)\u2014namely, CaBr<sub>2<\/sub> and KNO<sub>3<\/sub>. However, the solubility rules predict that these two substances would also be soluble, so no precipitate would form. Thus, we predict no reaction in this case.<\/li>\n<li>According to the solubility rules, both NaOH and FeCl<sub>2<\/sub> are expected to be soluble. If we assume that a double-replacement reaction may occur, we need to consider the possible products, which would be NaCl and Fe(OH)<sub>2<\/sub>. NaCl is soluble, but, according to the solubility rules, Fe(OH)<sub>2<\/sub> is not. Therefore, a reaction would occur, and Fe(OH)<sub>2<\/sub>(s) would precipitate out of solution. The balanced chemical equation is: 2NaOH(aq) + FeCl<sub>2<\/sub>(aq) \u2192 2NaCl(aq) + Fe(OH)<sub>2<\/sub>(s)<\/li>\n<\/ol>\n<h1>Test Yourself<\/h1>\n<p>Will a double-replacement equation occur? If so, identify the products.<\/p>\n<p style=\"text-align: center;\">Sr(NO<sub>3<\/sub>)<sub>2<\/sub> + KCl \u2192 ?<\/p>\n<h2>Answer<\/h2>\n<p>No reaction; all possible products are soluble.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Key Takeaways<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ul>\n<li>A single-replacement reaction replaces one element for another in a compound.<\/li>\n<li>The periodic table or an activity series can help predict whether single-replacement reactions occur.<\/li>\n<li>A double-replacement reaction exchanges the cations (or the anions) of two ionic compounds.<\/li>\n<li>A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate.<\/li>\n<li>Solubility rules are used to predict whether some double-replacement reactions will occur.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercises<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<h1>Questions<\/h1>\n<ol>\n<li>What are the general characteristics that help you recognize single-replacement reactions?<\/li>\n<li>What are the general characteristics that help you recognize double-replacement reactions?<\/li>\n<li>Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.\n<ol type=\"a\">\n<li>Zn + Fe(NO<sub>3<\/sub>)<sub>2<\/sub> \u2192 ?<\/li>\n<li>F<sub>2<\/sub> + FeI<sub>3<\/sub> \u2192 ?<\/li>\n<\/ol>\n<\/li>\n<li>Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.\n<ol type=\"a\">\n<li>Li + MgSO<sub>4<\/sub> \u2192 ?<\/li>\n<li>NaBr + Cl<sub>2<\/sub> \u2192 ?<\/li>\n<\/ol>\n<\/li>\n<li>Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.\n<ol type=\"a\">\n<li>Sn + H<sub>2<\/sub>SO<sub>4<\/sub> \u2192 ?<\/li>\n<li>Al + NiBr<sub>2<\/sub> \u2192 ?<\/li>\n<\/ol>\n<\/li>\n<li>Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.\n<ol type=\"a\">\n<li>Mg + HCl \u2192 ?<\/li>\n<li>HI + Br<sub>2<\/sub> \u2192 ?<\/li>\n<\/ol>\n<\/li>\n<li>Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.\n<ol type=\"a\">\n<li>FeCl<sub>2<\/sub> + Br<sub>2<\/sub> \u2192 ?<\/li>\n<li>Fe(NO<sub>3<\/sub>)<sub>3<\/sub> + Al \u2192 ?<\/li>\n<\/ol>\n<\/li>\n<li>Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.\n<ol type=\"a\">\n<li>Zn + Fe<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> \u2192 ?<\/li>\n<li>Ag + HNO<sub>3<\/sub> \u2192 ?<\/li>\n<\/ol>\n<\/li>\n<li>Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.\n<ol type=\"a\">\n<li>NaI + Cl<sub>2<\/sub> \u2192 ?<\/li>\n<li>AgCl + Au \u2192 ?<\/li>\n<\/ol>\n<\/li>\n<li>Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.\n<ol type=\"a\">\n<li>Pt + H<sub>3<\/sub>PO<sub>4<\/sub> \u2192 ?<\/li>\n<li>Li + H<sub>2<\/sub>O \u2192 ? (Hint: treat H<sub>2<\/sub>O as if it were composed of H<sup>+<\/sup> and OH<sup>\u2212<\/sup> ions.)<\/li>\n<\/ol>\n<\/li>\n<li>Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.\n<ol type=\"a\">\n<li>Zn(NO<sub>3<\/sub>)<sub>2<\/sub> + NaOH \u2192 ?<\/li>\n<li>HCl + Na<sub>2<\/sub>S \u2192 ?<\/li>\n<\/ol>\n<\/li>\n<li>Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.\n<ol type=\"a\">\n<li>Ca(C<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub>)<sub>2<\/sub> + HNO<sub>3<\/sub> \u2192 ?<\/li>\n<li>Na<sub>2<\/sub>CO<sub>3<\/sub> + Sr(NO<sub>2<\/sub>)<sub>2<\/sub> \u2192 ?<\/li>\n<\/ol>\n<\/li>\n<li>Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.\n<ol type=\"a\">\n<li>Pb(NO<sub>3<\/sub>)<sub>2<\/sub> + KBr \u2192 ?<\/li>\n<li>K<sub>2<\/sub>O + MgCO<sub>3<\/sub> \u2192 ?<\/li>\n<\/ol>\n<\/li>\n<li>Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.\n<ol type=\"a\">\n<li>Sn(OH)<sub>2<\/sub> + FeBr<sub>3<\/sub> \u2192 ?<\/li>\n<li>CsNO<sub>3<\/sub> + KCl \u2192 ?<\/li>\n<\/ol>\n<\/li>\n<li>Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.\n<ol type=\"a\">\n<li>Pb(NO<sub>3<\/sub>)<sub>2<\/sub> + KBr \u2192 ?<\/li>\n<li>K<sub>2<\/sub>O + Na<sub>2<\/sub>CO<sub>3<\/sub> \u2192 ?<\/li>\n<\/ol>\n<\/li>\n<li>Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.\n<ol type=\"a\">\n<li>Na<sub>2<\/sub>CO<sub>3<\/sub> + Sr(NO<sub>2<\/sub>)<sub>2<\/sub> \u2192 ?<\/li>\n<li>(NH<sub>4<\/sub>)<sub>2<\/sub>SO<sub>4<\/sub> + Ba(NO<sub>3<\/sub>)<sub>2<\/sub> \u2192 ?<\/li>\n<\/ol>\n<\/li>\n<li>Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.\n<ol type=\"a\">\n<li>K<sub>3<\/sub>PO<sub>4<\/sub> + SrCl<sub>2<\/sub> \u2192 ?<\/li>\n<li>NaOH + MgCl<sub>2<\/sub> \u2192 ?<\/li>\n<\/ol>\n<\/li>\n<li>Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.\n<ol type=\"a\">\n<li>KC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub> + Li<sub>2<\/sub>CO<sub>3<\/sub> \u2192 ?<\/li>\n<li>KOH + AgNO<sub>3<\/sub> \u2192 ?<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<h1>Answers<\/h1>\n<ol>\n<li>One element replaces another element in a compound.<\/li>\n<\/ol>\n<ol start=\"3\">\n<li>\n<ol type=\"a\">\n<li>Zn + Fe(NO<sub>3<\/sub>)<sub>2<\/sub> \u2192 Zn(NO<sub>3<\/sub>)<sub>2<\/sub> + Fe<\/li>\n<li>3F<sub>2<\/sub> + 2FeI<sub>3<\/sub> \u2192 3I<sub>2<\/sub> + 2FeF<sub>3<\/sub><\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<ol start=\"5\">\n<li>\n<ol type=\"a\">\n<li>Sn + H<sub>2<\/sub>SO<sub>4<\/sub> \u2192 SnSO<sub>4<\/sub> + H<sub>2<\/sub><\/li>\n<li>2Al + 3NiBr<sub>2<\/sub> \u2192 2AlBr<sub>3<\/sub> + 3Ni<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<ol start=\"7\">\n<li>\n<ol type=\"a\">\n<li>No reaction occurs.<\/li>\n<li>Fe(NO<sub>3<\/sub>)<sub>3<\/sub> + Al \u2192 Al(NO<sub>3<\/sub>)<sub>3<\/sub> + Fe<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<ol start=\"9\">\n<li>\n<ol type=\"a\">\n<li>2NaI + Cl<sub>2<\/sub> \u2192 2NaCl + I<sub>2<\/sub><\/li>\n<li>No reaction occurs.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<ol start=\"11\">\n<li>\n<ol type=\"a\">\n<li>Zn(NO<sub>3<\/sub>)<sub>2 <\/sub>+ 2NaOH \u2192 Zn(OH)<sub>2 <\/sub>+ 2NaNO<sub>3<\/sub><\/li>\n<li>2HCl + Na<sub>2<\/sub>S \u2192 2NaCl + H<sub>2<\/sub>S<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<ol start=\"13\">\n<li>\n<ol type=\"a\">\n<li>Pb(NO<sub>3<\/sub>)<sub>2<\/sub> + 2KBr \u2192 PbBr<sub>2<\/sub> + 2KNO<sub>3<\/sub><\/li>\n<li>K<sub>2<\/sub>O + MgCO<sub>3<\/sub> \u2192 K<sub>2<\/sub>CO<sub>3<\/sub> + MgO<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<ol start=\"15\">\n<li>\n<ol type=\"a\">\n<li>Pb(NO<sub>3<\/sub>)<sub>2<\/sub> + 2KBr \u2192 PbBr<sub>2<\/sub>(s) + 2KNO<sub>3<\/sub><\/li>\n<li>No reaction occurs.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<ol start=\"17\">\n<li>\n<ol type=\"a\">\n<li>2K<sub>3<\/sub>PO<sub>4<\/sub> + 3SrCl<sub>2<\/sub> \u2192 Sr<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>(s) + 6KCl<\/li>\n<li>2NaOH + MgCl<sub>2<\/sub> \u2192 2NaCl + Mg(OH)<sub>2<\/sub>(s)<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div id=\"ball-ch04_s02_n07\" class=\"key_takeaways editable block\">\n<h3>Media Attributions<\/h3>\n<p>Figure 4.1<\/p>\n<ul>\n<li>&#8220;Halogens on the Periodic Table&#8221; by David W. Ball \u00a9 <a href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/3.0\/\">CC BY-NC-SA (Attribution NonCommercial ShareAlike)<\/a><\/li>\n<\/ul>\n<p>Figure 4.2<\/p>\n<ul>\n<li><a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Copper_solution.jpg\">&#8220;Copper solution&#8221;<\/a> by Choij \u00a9 <a href=\"https:\/\/creativecommons.org\/publicdomain\/mark\/1.0\/\">Public Domain<\/a><\/li>\n<\/ul>\n<\/div>\n<div class=\"glossary\"><span class=\"screen-reader-text\" id=\"definition\">definition<\/span><template id=\"term_7324_8031\"><div class=\"glossary__definition\" role=\"dialog\" data-id=\"term_7324_8031\"><div tabindex=\"-1\"><p>A chemical reaction in which one element is substituted for another element in a compound.<\/p>\n<\/div><button><span aria-hidden=\"true\">&times;<\/span><span class=\"screen-reader-text\">Close definition<\/span><\/button><\/div><\/template><template id=\"term_7324_8044\"><div class=\"glossary__definition\" role=\"dialog\" data-id=\"term_7324_8044\"><div tabindex=\"-1\"><p>A list of elements that will replace elements below them in single-replacement reactions.<\/p>\n<\/div><button><span aria-hidden=\"true\">&times;<\/span><span class=\"screen-reader-text\">Close definition<\/span><\/button><\/div><\/template><template id=\"term_7324_8048\"><div class=\"glossary__definition\" role=\"dialog\" data-id=\"term_7324_8048\"><div tabindex=\"-1\"><p>A chemical reaction in which parts of two ionic compounds are exchanged.<\/p>\n<\/div><button><span aria-hidden=\"true\">&times;<\/span><span class=\"screen-reader-text\">Close definition<\/span><\/button><\/div><\/template><template id=\"term_7324_8046\"><div class=\"glossary__definition\" role=\"dialog\" data-id=\"term_7324_8046\"><div tabindex=\"-1\"><p>A chemical reaction in which two ionic compounds are dissolved in water and form a new ionic compound that does not dissolve.<\/p>\n<\/div><button><span aria-hidden=\"true\">&times;<\/span><span class=\"screen-reader-text\">Close definition<\/span><\/button><\/div><\/template><template id=\"term_7324_8047\"><div class=\"glossary__definition\" role=\"dialog\" data-id=\"term_7324_8047\"><div tabindex=\"-1\"><p>A solid that falls out of solution in a precipitation reaction.<\/p>\n<\/div><button><span aria-hidden=\"true\">&times;<\/span><span class=\"screen-reader-text\">Close definition<\/span><\/button><\/div><\/template><template id=\"term_7324_8049\"><div class=\"glossary__definition\" role=\"dialog\" data-id=\"term_7324_8049\"><div tabindex=\"-1\"><p>General statements that predict which ionic compounds dissolve and which do not.<\/p>\n<\/div><button><span aria-hidden=\"true\">&times;<\/span><span class=\"screen-reader-text\">Close definition<\/span><\/button><\/div><\/template><\/div>","protected":false},"author":90,"menu_order":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-7324","chapter","type-chapter","status-publish","hentry"],"part":7320,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7324","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":6,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7324\/revisions"}],"predecessor-version":[{"id":8776,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7324\/revisions\/8776"}],"part":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/parts\/7320"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7324\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/media?parent=7324"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=7324"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/contributor?post=7324"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/license?post=7324"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}