{"id":212172,"date":"2025-06-16T08:31:19","date_gmt":"2025-06-16T08:31:19","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=212172"},"modified":"2025-06-16T08:31:21","modified_gmt":"2025-06-16T08:31:21","slug":"balance-6-13-if-4-25-g-of-aluminum-reacts-with-16-3-g-of-chlorine-gas-to-produce-aluminum-chloride-according-to-the-following-unbalanced-equation-2-al-3cl2-2-aic-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/16\/balance-6-13-if-4-25-g-of-aluminum-reacts-with-16-3-g-of-chlorine-gas-to-produce-aluminum-chloride-according-to-the-following-unbalanced-equation-2-al-3cl2-2-aic-2\/","title":{"rendered":"Balance (6) 13. If 4.25 g of aluminum reacts with 16.3 g of chlorine gas to produce aluminum chloride according to the following unbalanced equation, 2 Al + 3Cl2 2 AIC"},"content":{"rendered":"\n

Balance (6) 13. If 4.25 g of aluminum reacts with 16.3 g of chlorine gas to produce aluminum chloride according to the following unbalanced equation, 2 Al + 3Cl2 2 AICI: a. Which chemical is the limiting reactant? (Show your calculations) Al-26.98×2 = 53.96 = 1.63% a-35.45 x6 = 212.3 = 6.11% 266.60 Al is the limiting rencant because it 4.25 will un act ofter 4 asg. Its the chemical 1621 with the smallest amung b. What is the theoretical yield (in grams) of aluminum chloride?<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/strong><\/mark><\/p>\n\n\n\n

To solve this problem, we must go through several steps: balancing the chemical equation<\/strong>, determining the limiting reactant<\/strong>, and then calculating the theoretical yield<\/strong> of aluminum chloride (AlCl\u2083<\/strong>).<\/p>\n\n\n\n

\n\n\n\n

Balanced Equation<\/strong><\/h3>\n\n\n\n

2Al+3Cl2\u21922AlCl32Al + 3Cl_2 \\rightarrow 2AlCl_32Al+3Cl2\u200b\u21922AlCl3\u200b<\/p>\n\n\n\n

\n\n\n\n

Step a: Determine the Limiting Reactant<\/strong><\/h3>\n\n\n\n

We are given:<\/p>\n\n\n\n

    \n
  • 4.25 g of Aluminum (Al)<\/strong><\/li>\n\n\n\n
  • 16.3 g of Chlorine gas (Cl\u2082)<\/strong><\/li>\n<\/ul>\n\n\n\n

    1. Molar Masses<\/strong>:<\/p>\n\n\n\n

      \n
    • Molar mass of Al<\/strong> = 26.98 g\/mol<\/li>\n\n\n\n
    • Molar mass of Cl\u2082<\/strong> = 70.90 g\/mol (because Cl = 35.45, and Cl\u2082 = 2 \u00d7 35.45)<\/li>\n<\/ul>\n\n\n\n

      2. Convert mass to moles<\/strong>:<\/p>\n\n\n\n

        \n
      • Moles of Al:<\/li>\n<\/ul>\n\n\n\n

        4.25\u2009g26.98\u2009g\/mol=0.1575\u2009mol\\frac{4.25 \\, \\text{g}}{26.98 \\, \\text{g\/mol}} = 0.1575 \\, \\text{mol}26.98g\/mol4.25g\u200b=0.1575mol<\/p>\n\n\n\n

          \n
        • Moles of Cl\u2082:<\/li>\n<\/ul>\n\n\n\n

          16.3\u2009g70.90\u2009g\/mol=0.2299\u2009mol\\frac{16.3 \\, \\text{g}}{70.90 \\, \\text{g\/mol}} = 0.2299 \\, \\text{mol}70.90g\/mol16.3g\u200b=0.2299mol<\/p>\n\n\n\n

          3. Determine limiting reactant using stoichiometry<\/strong>:
          From the balanced equation: 2 mol Al:3 mol Cl2\u21d20.15752=0.07875and0.22993=0.076632 \\text{ mol Al} : 3 \\text{ mol Cl}_2 \\Rightarrow \\frac{0.1575}{2} = 0.07875 \\quad \\text{and} \\quad \\frac{0.2299}{3} = 0.076632 mol Al:3 mol Cl2\u200b\u21d220.1575\u200b=0.07875and30.2299\u200b=0.07663<\/p>\n\n\n\n

          Since 0.07663 < 0.07875<\/strong>, chlorine gas (Cl\u2082)<\/strong> limits the reaction.<\/p>\n\n\n\n

          \n\n\n\n

          Step b: Theoretical Yield of AlCl\u2083<\/strong><\/h3>\n\n\n\n

          From the balanced equation: 3 mol Cl2\u21922 mol AlCl33 \\text{ mol Cl}_2 \\rightarrow 2 \\text{ mol AlCl}_33 mol Cl2\u200b\u21922 mol AlCl3\u200b<\/p>\n\n\n\n

          Using limiting reactant (Cl\u2082)<\/strong>:<\/p>\n\n\n\n

            \n
          • Moles of AlCl\u2083 produced:<\/li>\n<\/ul>\n\n\n\n

            0.2299\u2009mol Cl2\u00d72\u2009mol AlCl33\u2009mol Cl2=0.1533\u2009mol AlCl30.2299 \\, \\text{mol Cl}_2 \\times \\frac{2 \\, \\text{mol AlCl}_3}{3 \\, \\text{mol Cl}_2} = 0.1533 \\, \\text{mol AlCl}_30.2299mol Cl2\u200b\u00d73mol Cl2\u200b2mol AlCl3\u200b\u200b=0.1533mol AlCl3\u200b<\/p>\n\n\n\n

            Molar mass of AlCl\u2083<\/strong>:<\/p>\n\n\n\n

              \n
            • Al = 26.98<\/li>\n\n\n\n
            • Cl = 35.45 \u00d7 3 = 106.35<\/li>\n\n\n\n
            • Total = 133.33 g\/mol<\/li>\n<\/ul>\n\n\n\n

              Mass of AlCl\u2083 produced<\/strong>: 0.1533\u2009mol\u00d7133.33\u2009g\/mol=20.44\u2009g0.1533 \\, \\text{mol} \\times 133.33 \\, \\text{g\/mol} = \\boxed{20.44 \\, \\text{g}}0.1533mol\u00d7133.33g\/mol=20.44g\u200b<\/p>\n\n\n\n

              \n\n\n\n

              Final Answers<\/strong>:<\/h3>\n\n\n\n
                \n
              • Limiting Reactant<\/strong>: Chlorine gas (Cl\u2082<\/strong>)<\/li>\n\n\n\n
              • Theoretical Yield of AlCl\u2083<\/strong>: 20.44 g<\/strong><\/li>\n<\/ul>\n\n\n\n
                \n\n\n\n

                Explanation (300 words)<\/strong>:<\/h3>\n\n\n\n

                To find the theoretical yield in a chemical reaction, it’s essential first to identify the limiting reactant<\/strong>, which is the substance that runs out first and thus limits the amount of product that can form.<\/p>\n\n\n\n

                In this reaction, aluminum reacts with chlorine gas to form aluminum chloride according to the balanced equation: 2Al+3Cl2\u21922AlCl32Al + 3Cl_2 \\rightarrow 2AlCl_32Al+3Cl2\u200b\u21922AlCl3\u200b<\/p>\n\n\n\n

                We start by converting the given masses of aluminum (4.25 g) and chlorine gas (16.3 g) into moles using their molar masses. Aluminum has a molar mass of 26.98 g\/mol, and chlorine gas (Cl\u2082) has a molar mass of 70.90 g\/mol. This gives us 0.1575 mol of Al and 0.2299 mol of Cl\u2082.<\/p>\n\n\n\n

                Using the stoichiometric ratio from the balanced equation (2 mol Al : 3 mol Cl\u2082), we compare how much of each reactant would be needed. When we divide the moles of each reactant by its stoichiometric coefficient, we find that chlorine gas gives the smaller result. This means chlorine is the limiting reactant, because it will be consumed before all the aluminum reacts.<\/p>\n\n\n\n

                Next, we use the moles of the limiting reactant (Cl\u2082) to determine how much product (AlCl\u2083) can form. The molar ratio tells us that 3 mol of Cl\u2082 produces 2 mol of AlCl\u2083. Applying this ratio to 0.2299 mol of Cl\u2082, we calculate that 0.1533 mol of AlCl\u2083 can form. Finally, converting moles of AlCl\u2083 into grams using its molar mass (133.33 g\/mol), we find the theoretical yield is 20.44 grams<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"

                Balance (6) 13. If 4.25 g of aluminum reacts with 16.3 g of chlorine gas to produce aluminum chloride according to the following unbalanced equation, 2 Al + 3Cl2 2 AICI: a. Which chemical is the limiting reactant? (Show your calculations) Al-26.98×2 = 53.96 = 1.63% a-35.45 x6 = 212.3 = 6.11% 266.60 Al is […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-212172","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/212172","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=212172"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/212172\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=212172"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=212172"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=212172"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}