{"id":257694,"date":"2025-07-17T16:14:57","date_gmt":"2025-07-17T16:14:57","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=257694"},"modified":"2025-07-17T16:14:59","modified_gmt":"2025-07-17T16:14:59","slug":"calcium-carbonate-decomposes-into-calcium-oxide-and-carbon-dioxide-as-depicted-in-following-chemical-equation","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/17\/calcium-carbonate-decomposes-into-calcium-oxide-and-carbon-dioxide-as-depicted-in-following-chemical-equation\/","title":{"rendered":"Calcium carbonate decomposes into calcium oxide and carbon dioxide as depicted in following chemical equation."},"content":{"rendered":"\n

005 10.0 points Calcium carbonate decomposes into calcium oxide and carbon dioxide as depicted in following chemical equation. CaCO
(s)
CaO(s) + CO
(g) What is the total volume of gas produced by the decomposition of 25g of CaCO
at 3 atm and 315 K? Please answer in units of L. 1. 6.45L 2. 2.15 L 3. 18.5 L 4. 4.30 L 5. 5.60 L 6. 218 L<\/p>\n\n\n\n

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

To solve this, we need to determine the volume of CO\u2082 gas produced from the decomposition of calcium carbonate (CaCO\u2083). The balanced chemical equation is:<\/p>\n\n\n\n

CaCO\u2083(s) \u2192 CaO(s) + CO\u2082(g)<\/strong><\/p>\n\n\n\n

Step 1: Calculate the moles of CaCO\u2083<\/h3>\n\n\n\n

First, we need to find the molar mass of CaCO\u2083. Using the atomic masses:<\/p>\n\n\n\n

    \n
  • Ca: 40.08 g\/mol<\/li>\n\n\n\n
  • C: 12.01 g\/mol<\/li>\n\n\n\n
  • O: 16.00 g\/mol (3 oxygen atoms)<\/li>\n<\/ul>\n\n\n\n

    The molar mass of CaCO\u2083 is: Molar mass of CaCO\u2083=40.08+12.01+(3\u00d716.00)=100.09\u2009g\/mol\\text{Molar mass of CaCO\u2083} = 40.08 + 12.01 + (3 \\times 16.00) = 100.09 \\, \\text{g\/mol}Molar mass of CaCO\u2083=40.08+12.01+(3\u00d716.00)=100.09g\/mol<\/p>\n\n\n\n

    Next, we calculate the moles of CaCO\u2083 in 25 g: Moles of CaCO\u2083=25\u2009g100.09\u2009g\/mol=0.2498\u2009mol\\text{Moles of CaCO\u2083} = \\frac{25 \\, \\text{g}}{100.09 \\, \\text{g\/mol}} = 0.2498 \\, \\text{mol}Moles of CaCO\u2083=100.09g\/mol25g\u200b=0.2498mol<\/p>\n\n\n\n

    Step 2: Use the ideal gas law<\/h3>\n\n\n\n

    From the balanced equation, 1 mole of CaCO\u2083 produces 1 mole of CO\u2082. Therefore, 0.2498 moles of CaCO\u2083 will produce 0.2498 moles of CO\u2082.<\/p>\n\n\n\n

    To find the volume of CO\u2082 at 3 atm and 315 K, we use the ideal gas law: PV=nRTPV = nRTPV=nRT<\/p>\n\n\n\n

    Where:<\/p>\n\n\n\n

      \n
    • PPP = pressure = 3 atm<\/li>\n\n\n\n
    • VVV = volume (what we need to find)<\/li>\n\n\n\n
    • nnn = number of moles = 0.2498 mol<\/li>\n\n\n\n
    • RRR = ideal gas constant = 0.0821 L\u00b7atm\/(mol\u00b7K)<\/li>\n\n\n\n
    • TTT = temperature = 315 K<\/li>\n<\/ul>\n\n\n\n

      Rearrange the equation to solve for VVV: V=nRTPV = \\frac{nRT}{P}V=PnRT\u200b<\/p>\n\n\n\n

      Substitute the known values: V=(0.2498\u2009mol)\u00d7(0.0821\u2009L\\cdotpatm\/mol\\cdotpK)\u00d7(315\u2009K)3\u2009atm=6.4553=2.15\u2009LV = \\frac{(0.2498 \\, \\text{mol}) \\times (0.0821 \\, \\text{L\u00b7atm\/mol\u00b7K}) \\times (315 \\, \\text{K})}{3 \\, \\text{atm}} = \\frac{6.455}{3} = 2.15 \\, \\text{L}V=3atm(0.2498mol)\u00d7(0.0821L\\cdotpatm\/mol\\cdotpK)\u00d7(315K)\u200b=36.455\u200b=2.15L<\/p>\n\n\n\n

      Step 3: Conclusion<\/h3>\n\n\n\n

      The total volume of CO\u2082 gas produced is 2.15 L<\/strong>.<\/p>\n\n\n\n

      The correct answer is 2.15 L<\/strong> (Option 2).<\/p>\n\n\n\n

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

      005 10.0 points Calcium carbonate decomposes into calcium oxide and carbon dioxide as depicted in following chemical equation. CaCO(s)CaO(s) + CO(g) What is the total volume of gas produced by the decomposition of 25g of CaCOat 3 atm and 315 K? Please answer in units of L. 1. 6.45L 2. 2.15 L 3. 18.5 L […]<\/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-257694","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257694","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=257694"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257694\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257694"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257694"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257694"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}