{"id":241354,"date":"2025-07-03T17:09:05","date_gmt":"2025-07-03T17:09:05","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=241354"},"modified":"2025-07-03T17:09:08","modified_gmt":"2025-07-03T17:09:08","slug":"aluminum-reacts-with-bromine-to-form-aluminum-bromide-in-the-reaction","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/03\/aluminum-reacts-with-bromine-to-form-aluminum-bromide-in-the-reaction\/","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 (AlBr\u2083) will form, we need to use stoichiometry, which involves converting grams to moles, using the molar ratio from the balanced equation, and then converting the result back to grams.<\/p>\n\n\n\n

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

The balanced equation is:<\/p>\n\n\n\n

2Al+3Br2\u21922AlBr32Al + 3Br_2 \\rightarrow 2AlBr_32Al+3Br2\u200b\u21922AlBr3\u200b<\/p>\n\n\n\n

This shows that 2 moles of aluminum (Al) react with 3 moles of bromine (Br\u2082) to produce 2 moles of aluminum bromide (AlBr\u2083).<\/p>\n\n\n\n

Step 2: Convert grams to moles for both reactants<\/h3>\n\n\n\n
    \n
  • Molar mass of aluminum (Al)<\/strong> = 26.98 g\/mol<\/li>\n\n\n\n
  • Molar mass of bromine (Br\u2082)<\/strong> = 159.8 g\/mol (since the molar mass of Br is 79.9 g\/mol)<\/li>\n<\/ul>\n\n\n\n

    Moles of aluminum:<\/h4>\n\n\n\n

    Moles of Al=Mass of AlMolar mass of Al=15.79\u2009g26.98\u2009g\/mol=0.585\u2009mol Al\\text{Moles of Al} = \\frac{\\text{Mass of Al}}{\\text{Molar mass of Al}} = \\frac{15.79 \\, \\text{g}}{26.98 \\, \\text{g\/mol}} = 0.585 \\, \\text{mol Al}Moles of Al=Molar mass of AlMass of Al\u200b=26.98g\/mol15.79g\u200b=0.585mol Al<\/p>\n\n\n\n

    Moles of bromine:<\/h4>\n\n\n\n

    Moles of Br2=Mass of Br2Molar mass of Br2=55.61\u2009g159.8\u2009g\/mol=0.348\u2009mol Br2\\text{Moles of Br}_2 = \\frac{\\text{Mass of Br}_2}{\\text{Molar mass of Br}_2} = \\frac{55.61 \\, \\text{g}}{159.8 \\, \\text{g\/mol}} = 0.348 \\, \\text{mol Br}_2Moles of Br2\u200b=Molar mass of Br2\u200bMass of Br2\u200b\u200b=159.8g\/mol55.61g\u200b=0.348mol Br2\u200b<\/p>\n\n\n\n

    Step 3: Identify the limiting reactant<\/h3>\n\n\n\n

    The balanced equation tells us that the molar ratio between Al and Br\u2082 is 2:3. Let’s compare the ratio of the available moles of each reactant:0.585\u2009mol Al2=0.2925\\frac{0.585 \\, \\text{mol Al}}{2} = 0.292520.585mol Al\u200b=0.29250.348\u2009mol Br23=0.116\\frac{0.348 \\, \\text{mol Br}_2}{3} = 0.11630.348mol Br2\u200b\u200b=0.116<\/p>\n\n\n\n

    Since 0.116 mol of Br\u2082 is less than 0.2925 mol of Al, bromine (Br\u2082) is the limiting reactant.<\/p>\n\n\n\n

    Step 4: Use the limiting reactant to calculate the moles of AlBr\u2083 produced<\/h3>\n\n\n\n

    From the balanced equation, we know that 3 moles of Br\u2082 produce 2 moles of AlBr\u2083. Therefore, we can calculate the moles of AlBr\u2083 produced from 0.348 mol of Br\u2082:Moles of AlBr3=2\u2009mol AlBr33\u2009mol Br2\u00d70.348\u2009mol Br2=0.232\u2009mol AlBr3\\text{Moles of AlBr}_3 = \\frac{2 \\, \\text{mol AlBr}_3}{3 \\, \\text{mol Br}_2} \\times 0.348 \\, \\text{mol Br}_2 = 0.232 \\, \\text{mol AlBr}_3Moles of AlBr3\u200b=3mol Br2\u200b2mol AlBr3\u200b\u200b\u00d70.348mol Br2\u200b=0.232mol AlBr3\u200b<\/p>\n\n\n\n

    Step 5: Convert moles of AlBr\u2083 to grams<\/h3>\n\n\n\n

    The molar mass of AlBr\u2083 is:Molar mass of AlBr3=26.98\u2009g\/mol+3\u00d779.9\u2009g\/mol=266.68\u2009g\/mol\\text{Molar mass of AlBr}_3 = 26.98 \\, \\text{g\/mol} + 3 \\times 79.9 \\, \\text{g\/mol} = 266.68 \\, \\text{g\/mol}Molar mass of AlBr3\u200b=26.98g\/mol+3\u00d779.9g\/mol=266.68g\/mol<\/p>\n\n\n\n

    Now, we can calculate the mass of AlBr\u2083:Mass of AlBr3=Moles of AlBr3\u00d7Molar mass of AlBr3=0.232\u2009mol\u00d7266.68\u2009g\/mol=61.87\u2009g\\text{Mass of AlBr}_3 = \\text{Moles of AlBr}_3 \\times \\text{Molar mass of AlBr}_3 = 0.232 \\, \\text{mol} \\times 266.68 \\, \\text{g\/mol} = 61.87 \\, \\text{g}Mass of AlBr3\u200b=Moles of AlBr3\u200b\u00d7Molar mass of AlBr3\u200b=0.232mol\u00d7266.68g\/mol=61.87g<\/p>\n\n\n\n

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

    The mass of aluminum bromide (AlBr\u2083) that will form is 61.87 g<\/strong>.<\/p>\n\n\n\n

    Thus, the correct answer 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-241354","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/241354","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=241354"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/241354\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=241354"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=241354"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=241354"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}