{"id":36841,"date":"2025-06-25T09:47:03","date_gmt":"2025-06-25T09:47:03","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=36841"},"modified":"2025-06-25T09:47:05","modified_gmt":"2025-06-25T09:47:05","slug":"aluminum-reacts-with-chlorine-to-form-aluminum-chloride-2","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/aluminum-reacts-with-chlorine-to-form-aluminum-chloride-2\/","title":{"rendered":"Aluminum reacts with chlorine to form aluminum chloride."},"content":{"rendered":"\n<p>Aluminum reacts with chlorine to form aluminum chloride. The equation for this reaction is 2 Al(s) + 3 Cl2(g) -> 2 AlCl3(s). Match each situation to the corresponding percentage yield. Answer choices may be used only once. Use the molar mass values given here. Al = 26.98 g\/mol Cl2 = 70.90 g\/mol AlCl3 = 133.33 g\/mol &#8211; 20.00 g of chlorine was reacted with excess aluminum to produce 24.75 g of aluminum chloride. &#8211; 33.50 g of aluminum was reacted with excess chlorine to produce 164.5 g of aluminum chloride. &#8211; 145.0 g of aluminum was reacted with excess chlorine to produce 679.4 g of aluminum chloride. &#8211; 5.000 g of chlorine was reacted with excess aluminum to produce 6.000 g of aluminum chloride. Options: 1. 95.72% 2. 94.81% 3. 98.71% 4. 99.36% b) Hydrogen and nitrogen react to form ammonia. The chemical equation for this reaction is shown here: 3 H2(g) + N2(g) -> 2 NH3(g). Match each situation to the corresponding amount. Answer choices may be used only once. &#8211; Amount of H2 that will react with 8.2 mol of N2. &#8211; Amount of N2 that will produce 50.4 mol of NH3. &#8211; Amount of NH3 produced when 7.0 mol of N2 is consumed. &#8211; Amount of H2 that will produce 44.0 mol of NH3. &#8211; Amount of NH3 produced when 12.7 mol of H2 is consumed. &#8211; Amount of N2 that will react with 5.0 mol of H2. Options: 1. 8.47 mol 2. 66.0 mol 3. 25.2 mol 4. 24.6 mol 5. 1.67 mol 6. 14.0 mol<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part A: Aluminum reacts with chlorine to form aluminum chloride<\/strong><\/h3>\n\n\n\n<p><strong>Balanced equation:<\/strong><br>2 Al(s) + 3 Cl\u2082(g) \u2192 2 AlCl\u2083(s)<\/p>\n\n\n\n<p><strong>Molar masses:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Al = 26.98 g\/mol<\/li>\n\n\n\n<li>Cl\u2082 = 70.90 g\/mol<\/li>\n\n\n\n<li>AlCl\u2083 = 133.33 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>We\u2019ll use stoichiometry to find the theoretical yield of AlCl\u2083 and then use the formula:<\/p>\n\n\n\n<p><strong>% Yield = (Actual Yield \/ Theoretical Yield) \u00d7 100<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>20.00 g of Cl\u2082 produces 24.75 g of AlCl\u2083<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of Cl\u2082 = 20.00 g \u00f7 70.90 g\/mol = 0.282 mol<\/li>\n\n\n\n<li>From the equation, 3 mol Cl\u2082 produces 2 mol AlCl\u2083<\/li>\n\n\n\n<li>So, 0.282 mol Cl\u2082 produces:<br>(2\/3)\u00d70.282=0.188molAlCl3(2\/3) \u00d7 0.282 = 0.188 mol AlCl\u2083(2\/3)\u00d70.282=0.188molAlCl3\u200b<\/li>\n\n\n\n<li>Mass of AlCl\u2083 = 0.188 mol \u00d7 133.33 g\/mol = 25.06 g<\/li>\n\n\n\n<li>% Yield = (24.75 \u00f7 25.06) \u00d7 100 = <strong>98.71%<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li><strong>33.50 g Al produces 164.5 g AlCl\u2083<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of Al = 33.50 \u00f7 26.98 = 1.242 mol<\/li>\n\n\n\n<li>From the equation, 2 mol Al produces 2 mol AlCl\u2083 (1:1)<\/li>\n\n\n\n<li>So, 1.242 mol Al \u2192 1.242 mol AlCl\u2083<\/li>\n\n\n\n<li>Mass = 1.242 \u00d7 133.33 = 165.6 g<\/li>\n\n\n\n<li>% Yield = (164.5 \u00f7 165.6) \u00d7 100 = <strong>99.36%<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"3\" class=\"wp-block-list\">\n<li><strong>145.0 g Al produces 679.4 g AlCl\u2083<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of Al = 145.0 \u00f7 26.98 = 5.374 mol<\/li>\n\n\n\n<li>So, AlCl\u2083 = 5.374 mol<\/li>\n\n\n\n<li>Mass = 5.374 \u00d7 133.33 = 716.8 g<\/li>\n\n\n\n<li>% Yield = (679.4 \u00f7 716.8) \u00d7 100 = <strong>94.81%<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"4\" class=\"wp-block-list\">\n<li><strong>5.000 g Cl\u2082 produces 6.000 g AlCl\u2083<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles Cl\u2082 = 5.000 \u00f7 70.90 = 0.0705 mol<\/li>\n\n\n\n<li>Moles AlCl\u2083 = (2\/3) \u00d7 0.0705 = 0.047 mol<\/li>\n\n\n\n<li>Mass = 0.047 \u00d7 133.33 = 6.27 g<\/li>\n\n\n\n<li>% Yield = (6.000 \u00f7 6.27) \u00d7 100 = <strong>95.72%<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Matching:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>20.00 g Cl\u2082 \u2192 <strong>98.71%<\/strong><\/li>\n\n\n\n<li>33.50 g Al \u2192 <strong>99.36%<\/strong><\/li>\n\n\n\n<li>145.0 g Al \u2192 <strong>94.81%<\/strong><\/li>\n\n\n\n<li>5.000 g Cl\u2082 \u2192 <strong>95.72%<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part B: Hydrogen and nitrogen form ammonia<\/strong><\/h3>\n\n\n\n<p><strong>Balanced equation:<\/strong><br>3 H\u2082 + N\u2082 \u2192 2 NH\u2083<\/p>\n\n\n\n<p>Use stoichiometric ratios:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>3 mol H\u2082 per 2 mol NH\u2083<\/li>\n\n\n\n<li>1 mol N\u2082 per 2 mol NH\u2083<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Amount of H\u2082 to react with 8.2 mol N\u2082<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 N\u2082 : 3 H\u2082 \u2192 3 \u00d7 8.2 = <strong>24.6 mol<\/strong> \u2192 <strong>Option 4<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li><strong>Amount of N\u2082 to produce 50.4 mol NH\u2083<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>2 NH\u2083 : 1 N\u2082 \u2192 50.4 \u00f7 2 = <strong>25.2 mol<\/strong> \u2192 <strong>Option 3<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"3\" class=\"wp-block-list\">\n<li><strong>NH\u2083 produced from 7.0 mol N\u2082<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 N\u2082 \u2192 2 NH\u2083 \u2192 2 \u00d7 7.0 = <strong>14.0 mol<\/strong> \u2192 <strong>Option 6<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"4\" class=\"wp-block-list\">\n<li><strong>H\u2082 to produce 44.0 mol NH\u2083<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>2 NH\u2083 : 3 H\u2082 \u2192 (44.0 \u00d7 3) \u00f7 2 = <strong>66.0 mol<\/strong> \u2192 <strong>Option 2<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"5\" class=\"wp-block-list\">\n<li><strong>NH\u2083 produced from 12.7 mol H\u2082<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>3 H\u2082 \u2192 2 NH\u2083 \u2192 (12.7 \u00d7 2) \u00f7 3 = <strong>8.47 mol<\/strong> \u2192 <strong>Option 1<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"6\" class=\"wp-block-list\">\n<li><strong>N\u2082 to react with 5.0 mol H\u2082<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>3 H\u2082 : 1 N\u2082 \u2192 5.0 \u00f7 3 = <strong>1.67 mol<\/strong> \u2192 <strong>Option 5<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Matching:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>H\u2082 with 8.2 mol N\u2082 \u2192 <strong>24.6 mol<\/strong> \u2192 Option 4<\/li>\n\n\n\n<li>N\u2082 to make 50.4 mol NH\u2083 \u2192 <strong>25.2 mol<\/strong> \u2192 Option 3<\/li>\n\n\n\n<li>NH\u2083 from 7.0 mol N\u2082 \u2192 <strong>14.0 mol<\/strong> \u2192 Option 6<\/li>\n\n\n\n<li>H\u2082 for 44.0 mol NH\u2083 \u2192 <strong>66.0 mol<\/strong> \u2192 Option 2<\/li>\n\n\n\n<li>NH\u2083 from 12.7 mol H\u2082 \u2192 <strong>8.47 mol<\/strong> \u2192 Option 1<\/li>\n\n\n\n<li>N\u2082 for 5.0 mol H\u2082 \u2192 <strong>1.67 mol<\/strong> \u2192 Option 5<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-818.jpeg\" alt=\"\" class=\"wp-image-36847\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-818.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-818-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-818-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Aluminum reacts with chlorine to form aluminum chloride. The equation for this reaction is 2 Al(s) + 3 Cl2(g) -> 2 AlCl3(s). Match each situation to the corresponding percentage yield. Answer choices may be used only once. Use the molar mass values given here. Al = 26.98 g\/mol Cl2 = 70.90 g\/mol AlCl3 = 133.33 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-36841","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36841","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=36841"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36841\/revisions"}],"predecessor-version":[{"id":36848,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36841\/revisions\/36848"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=36841"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=36841"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=36841"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}