{"id":36700,"date":"2025-06-25T08:49:35","date_gmt":"2025-06-25T08:49:35","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=36700"},"modified":"2025-06-25T08:49:36","modified_gmt":"2025-06-25T08:49:36","slug":"aluminum-reacts-with-chlorine-to-form-aluminum-chloride","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/aluminum-reacts-with-chlorine-to-form-aluminum-chloride\/","title":{"rendered":"Aluminum reacts with chlorine to form aluminum chloride."},"content":{"rendered":"\n
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 – 20.00 g of chlorine was reacted with excess aluminum to produce 24.75 g of aluminum chloride. – 33.50 g of aluminum was reacted with excess chlorine to produce 164.5 g of aluminum chloride. – 145.0 g of aluminum was reacted with excess chlorine to produce 679.4 g of aluminum chloride. – 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. – Amount of H2 that will react with 8.2 mol of N2. – Amount of N2 that will produce 50.4 mol of NH3. – Amount of NH3 produced when 7.0 mol of N2 is consumed. – Amount of H2 that will produce 44.0 mol of NH3. – Amount of NH3 produced when 12.7 mol of H2 is consumed. – 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
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Part A: Aluminum and Chlorine Reaction<\/strong><\/h3>\n\n\n\n
We use stoichiometry to determine theoretical yields and then compare with actual yields to calculate percent yield using the formula: Percent yield = (Actual yield \/ Theoretical yield) \u00d7 100<\/strong><\/p>\n\n\n\n\n\n\n\n
1.<\/strong> 20.00 g of chlorine reacted with excess aluminum to produce 24.75 g of aluminum chloride.<\/em><\/p>\n\n\n\n
Moles of Cl\u2082 = 20.00 g \u00f7 70.90 g\/mol = 0.2821 mol From stoichiometry: 3 mol Cl\u2082 produce 2 mol AlCl\u2083 So 0.2821 mol Cl\u2082 produces: (2\/3) \u00d7 0.2821 = 0.1881 mol AlCl\u2083 Mass = 0.1881 mol \u00d7 133.33 g\/mol = 25.07 g<\/strong> (theoretical yield) % yield = (24.75 \u00f7 25.07) \u00d7 100 = 98.71%<\/strong> \u2192 Match: 3. 98.71%<\/strong><\/p>\n\n\n\n\n\n\n\n
2.<\/strong> 33.50 g of aluminum reacted with excess chlorine to produce 164.5 g of aluminum chloride.<\/em><\/p>\n\n\n\n
Moles of Al = 33.50 \u00f7 26.98 = 1.2417 mol From stoichiometry: 2 mol Al produce 2 mol AlCl\u2083 \u2192 1:1 So, 1.2417 mol Al \u2192 1.2417 mol AlCl\u2083 Mass = 1.2417 \u00d7 133.33 = 165.55 g<\/strong> (theoretical yield) % yield = (164.5 \u00f7 165.55) \u00d7 100 = 99.36%<\/strong> \u2192 Match: 4. 99.36%<\/strong><\/p>\n\n\n\n\n\n\n\n
3.<\/strong> 145.0 g of aluminum reacted with excess chlorine to produce 679.4 g of aluminum chloride.<\/em><\/p>\n\n\n\n
Moles of Al = 145.0 \u00f7 26.98 = 5.3733 mol So 5.3733 mol Al \u2192 5.3733 mol AlCl\u2083 Mass = 5.3733 \u00d7 133.33 = 716.37 g<\/strong> % yield = (679.4 \u00f7 716.37) \u00d7 100 = 94.81%<\/strong> \u2192 Match: 2. 94.81%<\/strong><\/p>\n\n\n\n\n\n\n\n
4.<\/strong> 5.000 g of chlorine reacted with excess aluminum to produce 6.000 g of aluminum chloride.<\/em><\/p>\n\n\n\n
Use mole ratios from the equation to match:<\/p>\n\n\n\n
\n\n\n\n
1. Amount of H\u2082 that will react with 8.2 mol of N\u2082<\/strong> 3 mol H\u2082 per 1 mol N\u2082 \u2192 3 \u00d7 8.2 = 24.6 mol<\/strong> \u2192 Match: 4. 24.6 mol<\/strong><\/p>\n\n\n\n\n\n\n\n
2. Amount of N\u2082 that will produce 50.4 mol of NH\u2083<\/strong> From 1 mol N\u2082 \u2192 2 mol NH\u2083 N\u2082 = 50.4 \u00f7 2 = 25.2 mol<\/strong> \u2192 Match: 3. 25.2 mol<\/strong><\/p>\n\n\n\n\n\n\n\n
3. Amount of NH\u2083 produced when 7.0 mol of N\u2082 is consumed<\/strong> 2 mol NH\u2083 per 1 mol N\u2082 \u2192 2 \u00d7 7.0 = 14.0 mol<\/strong> \u2192 Match: 6. 14.0 mol<\/strong><\/p>\n\n\n\n\n\n\n\n
4. Amount of H\u2082 that will produce 44.0 mol of NH\u2083<\/strong> From 2 mol NH\u2083 needs 3 mol H\u2082 H\u2082 = 44.0 \u00d7 (3\/2) = 66.0 mol<\/strong> \u2192 Match: 2. 66.0 mol<\/strong><\/p>\n\n\n\n\n\n\n\n
5. Amount of NH\u2083 produced when 12.7 mol of H\u2082 is consumed<\/strong> From 3 mol H\u2082 \u2192 2 mol NH\u2083 NH\u2083 = 12.7 \u00d7 (2\/3) = 8.47 mol<\/strong> \u2192 Match: 1. 8.47 mol<\/strong><\/p>\n\n\n\n\n\n\n\n
6. Amount of N\u2082 that will react with 5.0 mol of H\u2082<\/strong> From 3 mol H\u2082 \u2192 1 mol N\u2082 N\u2082 = 5.0 \u00f7 3 = 1.67 mol<\/strong> \u2192 Match: 5. 1.67 mol<\/strong><\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
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 […]<\/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-36700","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36700","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=36700"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36700\/revisions"}],"predecessor-version":[{"id":36709,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36700\/revisions\/36709"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=36700"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=36700"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=36700"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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