{"id":241432,"date":"2025-07-03T17:53:27","date_gmt":"2025-07-03T17:53:27","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=241432"},"modified":"2025-07-03T17:53:29","modified_gmt":"2025-07-03T17:53:29","slug":"aluminum-reacts-with-bromine-to-form-aluminum-bromide-in-the-reaction-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/03\/aluminum-reacts-with-bromine-to-form-aluminum-bromide-in-the-reaction-2\/","title":{"rendered":"Aluminum reacts with bromine to form aluminum bromide in the reaction"},"content":{"rendered":"\n

Aluminum reacts with bromine to form aluminum bromide in the reaction: 2Al + 3Br2 \u00e2\u2020\u2019 2AlBr3 If 15.79 g of aluminum reacts with 55.61 g of bromine, how many grams of aluminum bromide will form? 1. 78.01 g 2. 156.02 g 3. 123.75 g 4. 30.94 g 5. 61.87 g?<\/p>\n\n\n\n

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

To determine how many grams of aluminum bromide (AlBr3) will form, we need to follow these steps:<\/p>\n\n\n\n

Step 1: Write the balanced chemical equation.<\/h3>\n\n\n\n

The reaction between aluminum (Al) and bromine (Br2) to form aluminum bromide is:<\/p>\n\n\n\n

2Al + 3Br2 \u2192 2AlBr3<\/strong><\/p>\n\n\n\n

Step 2: Find the molar masses of aluminum, bromine, and aluminum bromide.<\/h3>\n\n\n\n
    \n
  • Molar mass of Al = 26.98 g\/mol<\/li>\n\n\n\n
  • Molar mass of Br = 79.90 g\/mol<\/li>\n\n\n\n
  • Molar mass of AlBr3 = (26.98 g\/mol) + 3 \u00d7 (79.90 g\/mol) = 266.68 g\/mol<\/li>\n<\/ul>\n\n\n\n

    Step 3: Convert the masses of aluminum and bromine to moles.<\/h3>\n\n\n\n

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

      \n
    • Mass of Al = 15.79 g<\/li>\n\n\n\n
    • Mass of Br2 = 55.61 g<\/li>\n<\/ul>\n\n\n\n

      For aluminum:moles of Al=15.79\u2009g26.98\u2009g\/mol=0.585\u2009mol\\text{moles of Al} = \\frac{15.79 \\, \\text{g}}{26.98 \\, \\text{g\/mol}} = 0.585 \\, \\text{mol}moles of Al=26.98g\/mol15.79g\u200b=0.585mol<\/p>\n\n\n\n

      For bromine:moles of Br2=55.61\u2009g159.80\u2009g\/mol=0.348\u2009mol\\text{moles of Br2} = \\frac{55.61 \\, \\text{g}}{159.80 \\, \\text{g\/mol}} = 0.348 \\, \\text{mol}moles of Br2=159.80g\/mol55.61g\u200b=0.348mol<\/p>\n\n\n\n

      Step 4: Determine the limiting reactant.<\/h3>\n\n\n\n

      From the balanced equation, we know:<\/p>\n\n\n\n

        \n
      • 2 moles of Al react with 3 moles of Br2.<\/li>\n<\/ul>\n\n\n\n

        Thus, the mole ratio of Al to Br2 is 2:3. To find the limiting reactant, we need to check which reactant will run out first.<\/p>\n\n\n\n

        For 0.585 moles of Al, we need:moles of Br2=32\u00d70.585=0.878\u2009mol\\text{moles of Br2} = \\frac{3}{2} \\times 0.585 = 0.878 \\, \\text{mol}moles of Br2=23\u200b\u00d70.585=0.878mol<\/p>\n\n\n\n

        But we only have 0.348 moles of Br2, which is less than 0.878 mol, so bromine (Br2) is the limiting reactant<\/strong>.<\/p>\n\n\n\n

        Step 5: Calculate the amount of AlBr3 produced.<\/h3>\n\n\n\n

        From the balanced equation:<\/p>\n\n\n\n

          \n
        • 3 moles of Br2 produce 2 moles of AlBr3.<\/li>\n<\/ul>\n\n\n\n

          So, 0.348 moles of Br2 will produce:moles of AlBr3=23\u00d70.348=0.232\u2009mol\\text{moles of AlBr3} = \\frac{2}{3} \\times 0.348 = 0.232 \\, \\text{mol}moles of AlBr3=32\u200b\u00d70.348=0.232mol<\/p>\n\n\n\n

          Now, we can calculate the mass of AlBr3 produced:mass of AlBr3=0.232\u2009mol\u00d7266.68\u2009g\/mol=61.87\u2009g\\text{mass of AlBr3} = 0.232 \\, \\text{mol} \\times 266.68 \\, \\text{g\/mol} = 61.87 \\, \\text{g}mass of AlBr3=0.232mol\u00d7266.68g\/mol=61.87g<\/p>\n\n\n\n

          Final Answer:<\/h3>\n\n\n\n

          61.87 g<\/strong> of aluminum bromide will form. Therefore, the correct option is:
          5. 61.87 g<\/strong>.<\/p>\n\n\n\n

          \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

          Aluminum reacts with bromine to form aluminum bromide in the reaction: 2Al + 3Br2 \u00e2\u2020\u2019 2AlBr3 If 15.79 g of aluminum reacts with 55.61 g of bromine, how many grams of aluminum bromide will form? 1. 78.01 g 2. 156.02 g 3. 123.75 g 4. 30.94 g 5. 61.87 g? The Correct Answer and Explanation […]<\/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-241432","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/241432","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=241432"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/241432\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=241432"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=241432"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=241432"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}