{"id":226,"date":"2016-01-11T19:59:47","date_gmt":"2016-01-11T19:59:47","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/end-of-chapter-material-21\/"},"modified":"2020-05-01T20:36:18","modified_gmt":"2020-05-01T20:36:18","slug":"end-of-chapter-5-material","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/end-of-chapter-5-material\/","title":{"raw":"End-of-Chapter Material","rendered":"End-of-Chapter Material"},"content":{"raw":"<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Additional Exercises<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>How many molecules of O<sub>2<\/sub> will react with 6.022 \u00d7 10<sup>23<\/sup> molecules of H<sub>2<\/sub> to make water? The reaction is 2H<sub>2<\/sub>(g) +\u00a0O<sub>2<\/sub>(g) \u2192\u00a02H<sub>2<\/sub>O(\u2113).<\/li>\r\n \t<li>How many molecules of H<sub>2<\/sub> will react with 6.022 \u00d7 10<sup>23<\/sup> molecules of N<sub>2<\/sub> to make ammonia? The reaction is N<sub>2<\/sub>(g) +\u00a03H<sub>2<\/sub>(g) \u2192\u00a02NH<sub>3<\/sub>(g).<\/li>\r\n \t<li>How many moles are present in 6.411 kg of CO<sub>2<\/sub>? How many molecules is this?<\/li>\r\n \t<li>How many moles are present in 2.998 mg of SCl<sub>4<\/sub>? How many molecules is this?<\/li>\r\n \t<li>What is the mass in milligrams of 7.22 \u00d7 10<sup>20<\/sup> molecules of CO<sub>2<\/sub>?<\/li>\r\n \t<li>What is the mass in kilograms of 3.408 \u00d7 10<sup>25<\/sup> molecules of SiS<sub>2<\/sub>?<\/li>\r\n \t<li>What is the mass in grams of 1 molecule of H<sub>2<\/sub>O?<\/li>\r\n \t<li>What is the mass in grams of 1 atom of Al?<\/li>\r\n \t<li>What is the volume of 3.44 mol of Ga if the density of Ga is 6.08 g\/mL?<\/li>\r\n \t<li>What is the volume of 0.662 mol of He if the density of He is 0.1785 g\/L?<\/li>\r\n \t<li>For the chemical reaction 2C<sub>4<\/sub>H<sub>10<\/sub>(g) +\u00a013O<sub>2<\/sub>(g) \u2192\u00a08CO<sub>2<\/sub>(g) +\u00a010H<sub>2<\/sub>O(\u2113), assume that 13.4 g of C<sub>4<\/sub>H<sub>10<\/sub> reacts completely to products. The density of CO<sub>2<\/sub> is 1.96 g\/L. What volume in litres of CO<sub>2<\/sub> is produced?<\/li>\r\n \t<li>For the chemical reaction 2GaCl<sub>3<\/sub>(s) +\u00a03H<sub>2<\/sub>(g) \u2192\u00a02Ga(\u2113) +\u00a06HCl(g), if 223 g of GaCl<sub>3<\/sub> reacts completely to products and the density of Ga is 6.08 g\/mL, what volume in millilitres of Ga is produced?<\/li>\r\n \t<li>Calculate the mass of each product when 100.0 g of CuCl react according to the reaction 2CuCl(aq) \u2192 CuCl<sub>2<\/sub>(aq) +\u00a0Cu(s). What do you notice about the sum of the masses of the products? What concept is being illustrated here?<\/li>\r\n \t<li>Calculate the mass of each product when 500.0 g of SnCl<sub>2<\/sub> react according to the reaction 2SnCl<sub>2<\/sub>(aq) \u2192\u00a0SnCl<sub>4<\/sub>(aq) +\u00a0Sn(s). What do you notice about the sum of the masses of the products? What concept is being illustrated here?<\/li>\r\n \t<li>What mass of CO<sub>2<\/sub> is produced from the combustion of 1 gal of gasoline? The chemical formula of gasoline can be approximated as C<sub>8<\/sub>H<sub>18<\/sub>. Assume that there are 2,801 g of gasoline per gallon.<\/li>\r\n \t<li>What mass of H<sub>2<\/sub>O is produced from the combustion of 1 gal of gasoline? The chemical formula of gasoline can be approximated as C<sub>8<\/sub>H<sub>18<\/sub>. Assume that there are 2,801 g of gasoline per gallon.<\/li>\r\n \t<li>A chemical reaction has a theoretical yield of 19.98 g and a percent yield of 88.40%. What is the actual yield?<\/li>\r\n \t<li>A chemical reaction has an actual yield of 19.98 g and a percent yield of 88.40%. What is the theoretical yield?<\/li>\r\n \t<li>Given the initial amounts listed, what is the limiting reagent, and how much of the other reactants are in excess?\r\n35.0 g P<sub>4<\/sub> + 12.7 g 3NaOH + 9.33 g 3H<sub>2<\/sub>O \u2192 2Na<sub>2<\/sub>HPO<sub>4<\/sub> + PH<sub>3<\/sub><\/li>\r\n \t<li>Given the initial amounts listed, what is the limiting reagent, and how much of the other reactants are in excess?\r\n46.3 g 2NaCrO<sub>2<\/sub> + 88.2 g 3NaBrO<sub>4<\/sub> + 32.5 g 2NaOH \u2192 3NaBrO<sub>3<\/sub> + 2Na<sub>2<\/sub>CrO<sub>4<\/sub> + H<sub>2<\/sub>O<\/li>\r\n \t<li>Verify that it does not matter which product you use to predict the limiting reagent by using both products in this combustion reaction to determine the limiting reagent and the amount of the reactant in excess. Initial amounts of each reactant are given.\r\n26.3 g C<sub>3<\/sub>H<sub>8<\/sub> + 21.8 g 5O<sub>2<\/sub> \u2192 3CO<sub>2<\/sub>(g) + 4H<sub>2<\/sub>O(\u2113)<\/li>\r\n \t<li>Just in case you suspect Exercise 21 is rigged, do it for another chemical reaction and verify that it does not matter which product you use to predict the limiting reagent by using both products in this combustion reaction to determine the limiting reagent and the amount of the reactant in excess. Initial amounts of each reactant are given.\r\n35.0 g 2P<sub>4<\/sub> + 12.7 g 6NaOH + 9.33 g 6H<sub>2<\/sub>O \u2192 3Na<sub>2<\/sub>HPO<sub>4<\/sub> + 5PH<sub>3<\/sub><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Answers<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>1.2044 \u00d7 10<sup>24<\/sup> molecules<\/li>\r\n<\/ol>\r\n<ol start=\"3\">\r\n \t<li>145.7 mol; 8.77 \u00d7 10<sup>25<\/sup> molecules<\/li>\r\n<\/ol>\r\n<ol start=\"5\">\r\n \t<li>52.8 mg<\/li>\r\n<\/ol>\r\n<ol start=\"7\">\r\n \t<li>2.99 \u00d7 10<sup>\u221223<\/sup> g<\/li>\r\n<\/ol>\r\n<ol start=\"9\">\r\n \t<li>39.4 mL<\/li>\r\n<\/ol>\r\n<ol start=\"11\">\r\n \t<li>20.7 L<\/li>\r\n<\/ol>\r\n<ol start=\"13\">\r\n \t<li>67.91 g of CuCl<sub>2<\/sub>; 32.09 g of Cu. The two masses add to 100.0 g, the initial amount of starting material, demonstrating the law of conservation of matter.<\/li>\r\n<\/ol>\r\n<ol start=\"15\">\r\n \t<li>8,632 g<\/li>\r\n<\/ol>\r\n<ol start=\"17\">\r\n \t<li>17.66 g<\/li>\r\n<\/ol>\r\n<ol start=\"19\">\r\n \t<li>The limiting reagent is NaOH; 21.9 g of P<sub>4<\/sub> and 3.61 g of H<sub>2<\/sub>O are left over.<\/li>\r\n<\/ol>\r\n<ol start=\"21\">\r\n \t<li>Both products predict that O<sub>2<\/sub> is the limiting reagent; 20.3 g of C<sub>3<\/sub>H<sub>8<\/sub> are left over.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Additional Exercises<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>How many molecules of O<sub>2<\/sub> will react with 6.022 \u00d7 10<sup>23<\/sup> molecules of H<sub>2<\/sub> to make water? The reaction is 2H<sub>2<\/sub>(g) +\u00a0O<sub>2<\/sub>(g) \u2192\u00a02H<sub>2<\/sub>O(\u2113).<\/li>\n<li>How many molecules of H<sub>2<\/sub> will react with 6.022 \u00d7 10<sup>23<\/sup> molecules of N<sub>2<\/sub> to make ammonia? The reaction is N<sub>2<\/sub>(g) +\u00a03H<sub>2<\/sub>(g) \u2192\u00a02NH<sub>3<\/sub>(g).<\/li>\n<li>How many moles are present in 6.411 kg of CO<sub>2<\/sub>? How many molecules is this?<\/li>\n<li>How many moles are present in 2.998 mg of SCl<sub>4<\/sub>? How many molecules is this?<\/li>\n<li>What is the mass in milligrams of 7.22 \u00d7 10<sup>20<\/sup> molecules of CO<sub>2<\/sub>?<\/li>\n<li>What is the mass in kilograms of 3.408 \u00d7 10<sup>25<\/sup> molecules of SiS<sub>2<\/sub>?<\/li>\n<li>What is the mass in grams of 1 molecule of H<sub>2<\/sub>O?<\/li>\n<li>What is the mass in grams of 1 atom of Al?<\/li>\n<li>What is the volume of 3.44 mol of Ga if the density of Ga is 6.08 g\/mL?<\/li>\n<li>What is the volume of 0.662 mol of He if the density of He is 0.1785 g\/L?<\/li>\n<li>For the chemical reaction 2C<sub>4<\/sub>H<sub>10<\/sub>(g) +\u00a013O<sub>2<\/sub>(g) \u2192\u00a08CO<sub>2<\/sub>(g) +\u00a010H<sub>2<\/sub>O(\u2113), assume that 13.4 g of C<sub>4<\/sub>H<sub>10<\/sub> reacts completely to products. The density of CO<sub>2<\/sub> is 1.96 g\/L. What volume in litres of CO<sub>2<\/sub> is produced?<\/li>\n<li>For the chemical reaction 2GaCl<sub>3<\/sub>(s) +\u00a03H<sub>2<\/sub>(g) \u2192\u00a02Ga(\u2113) +\u00a06HCl(g), if 223 g of GaCl<sub>3<\/sub> reacts completely to products and the density of Ga is 6.08 g\/mL, what volume in millilitres of Ga is produced?<\/li>\n<li>Calculate the mass of each product when 100.0 g of CuCl react according to the reaction 2CuCl(aq) \u2192 CuCl<sub>2<\/sub>(aq) +\u00a0Cu(s). What do you notice about the sum of the masses of the products? What concept is being illustrated here?<\/li>\n<li>Calculate the mass of each product when 500.0 g of SnCl<sub>2<\/sub> react according to the reaction 2SnCl<sub>2<\/sub>(aq) \u2192\u00a0SnCl<sub>4<\/sub>(aq) +\u00a0Sn(s). What do you notice about the sum of the masses of the products? What concept is being illustrated here?<\/li>\n<li>What mass of CO<sub>2<\/sub> is produced from the combustion of 1 gal of gasoline? The chemical formula of gasoline can be approximated as C<sub>8<\/sub>H<sub>18<\/sub>. Assume that there are 2,801 g of gasoline per gallon.<\/li>\n<li>What mass of H<sub>2<\/sub>O is produced from the combustion of 1 gal of gasoline? The chemical formula of gasoline can be approximated as C<sub>8<\/sub>H<sub>18<\/sub>. Assume that there are 2,801 g of gasoline per gallon.<\/li>\n<li>A chemical reaction has a theoretical yield of 19.98 g and a percent yield of 88.40%. What is the actual yield?<\/li>\n<li>A chemical reaction has an actual yield of 19.98 g and a percent yield of 88.40%. What is the theoretical yield?<\/li>\n<li>Given the initial amounts listed, what is the limiting reagent, and how much of the other reactants are in excess?<br \/>\n35.0 g P<sub>4<\/sub> + 12.7 g 3NaOH + 9.33 g 3H<sub>2<\/sub>O \u2192 2Na<sub>2<\/sub>HPO<sub>4<\/sub> + PH<sub>3<\/sub><\/li>\n<li>Given the initial amounts listed, what is the limiting reagent, and how much of the other reactants are in excess?<br \/>\n46.3 g 2NaCrO<sub>2<\/sub> + 88.2 g 3NaBrO<sub>4<\/sub> + 32.5 g 2NaOH \u2192 3NaBrO<sub>3<\/sub> + 2Na<sub>2<\/sub>CrO<sub>4<\/sub> + H<sub>2<\/sub>O<\/li>\n<li>Verify that it does not matter which product you use to predict the limiting reagent by using both products in this combustion reaction to determine the limiting reagent and the amount of the reactant in excess. Initial amounts of each reactant are given.<br \/>\n26.3 g C<sub>3<\/sub>H<sub>8<\/sub> + 21.8 g 5O<sub>2<\/sub> \u2192 3CO<sub>2<\/sub>(g) + 4H<sub>2<\/sub>O(\u2113)<\/li>\n<li>Just in case you suspect Exercise 21 is rigged, do it for another chemical reaction and verify that it does not matter which product you use to predict the limiting reagent by using both products in this combustion reaction to determine the limiting reagent and the amount of the reactant in excess. Initial amounts of each reactant are given.<br \/>\n35.0 g 2P<sub>4<\/sub> + 12.7 g 6NaOH + 9.33 g 6H<sub>2<\/sub>O \u2192 3Na<sub>2<\/sub>HPO<sub>4<\/sub> + 5PH<sub>3<\/sub><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Answers<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>1.2044 \u00d7 10<sup>24<\/sup> molecules<\/li>\n<\/ol>\n<ol start=\"3\">\n<li>145.7 mol; 8.77 \u00d7 10<sup>25<\/sup> molecules<\/li>\n<\/ol>\n<ol start=\"5\">\n<li>52.8 mg<\/li>\n<\/ol>\n<ol start=\"7\">\n<li>2.99 \u00d7 10<sup>\u221223<\/sup> g<\/li>\n<\/ol>\n<ol start=\"9\">\n<li>39.4 mL<\/li>\n<\/ol>\n<ol start=\"11\">\n<li>20.7 L<\/li>\n<\/ol>\n<ol start=\"13\">\n<li>67.91 g of CuCl<sub>2<\/sub>; 32.09 g of Cu. The two masses add to 100.0 g, the initial amount of starting material, demonstrating the law of conservation of matter.<\/li>\n<\/ol>\n<ol start=\"15\">\n<li>8,632 g<\/li>\n<\/ol>\n<ol start=\"17\">\n<li>17.66 g<\/li>\n<\/ol>\n<ol start=\"19\">\n<li>The limiting reagent is NaOH; 21.9 g of P<sub>4<\/sub> and 3.61 g of H<sub>2<\/sub>O are left over.<\/li>\n<\/ol>\n<ol start=\"21\">\n<li>Both products predict that O<sub>2<\/sub> is the limiting reagent; 20.3 g of C<sub>3<\/sub>H<sub>8<\/sub> are left over.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n","protected":false},"author":124,"menu_order":7,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-226","chapter","type-chapter","status-publish","hentry"],"part":156,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters\/226","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/users\/124"}],"version-history":[{"count":4,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters\/226\/revisions"}],"predecessor-version":[{"id":1540,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters\/226\/revisions\/1540"}],"part":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/parts\/156"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters\/226\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/media?parent=226"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapter-type?post=226"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/contributor?post=226"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/license?post=226"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}