{"id":257713,"date":"2025-07-17T16:30:32","date_gmt":"2025-07-17T16:30:32","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=257713"},"modified":"2025-07-17T16:30:35","modified_gmt":"2025-07-17T16:30:35","slug":"calcium-oxide-can-be-used-to-scrub-carbon-dioxide-from-air-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/17\/calcium-oxide-can-be-used-to-scrub-carbon-dioxide-from-air-2\/","title":{"rendered":"Calcium oxide can be used to “scrub” carbon dioxide from air."},"content":{"rendered":"\n

Calcium oxide can be used to “scrub” carbon dioxide from air. CaO (s) + CO2 (g) –> CaCO3 (s) What mass of CO2 could be absorbed by 1.85 g of ? Mass = g CO2 What volume would this CO2 occupy at STP? Volume = L CO2<\/p>\n\n\n\n

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

Mass of CO2 Absorbed by 1.85 g of CaO:<\/h3>\n\n\n\n

We are given the reaction:
CaO (s)+CO2(g)\u2192CaCO3(s)\\text{CaO (s)} + \\text{CO}_2 \\text{(g)} \\to \\text{CaCO}_3 \\text{(s)}CaO (s)+CO2\u200b(g)\u2192CaCO3\u200b(s)<\/p>\n\n\n\n

From the balanced equation, we know that 1 mole of CaO reacts with 1 mole of CO2 to produce 1 mole of CaCO3. So, the molar ratio between CaO and CO2 is 1:1.<\/p>\n\n\n\n

Step 1: Molar Mass of CaO<\/h4>\n\n\n\n

The molar mass of calcium oxide (CaO) is:<\/p>\n\n\n\n

    \n
  • Ca=40.08\u2009g\/mol\\text{Ca} = 40.08 \\, \\text{g\/mol}Ca=40.08g\/mol<\/li>\n\n\n\n
  • O=16.00\u2009g\/mol\\text{O} = 16.00 \\, \\text{g\/mol}O=16.00g\/mol<\/li>\n<\/ul>\n\n\n\n

    Thus, the molar mass of CaO is:
    40.08+16.00=56.08\u2009g\/mol40.08 + 16.00 = 56.08 \\, \\text{g\/mol}40.08+16.00=56.08g\/mol<\/p>\n\n\n\n

    Step 2: Moles of CaO in 1.85 g<\/h4>\n\n\n\n

    The number of moles of CaO in 1.85 g is calculated as:moles of CaO=mass of CaOmolar mass of CaO=1.85\u2009g56.08\u2009g\/mol\u22480.03296\u2009mol\\text{moles of CaO} = \\frac{\\text{mass of CaO}}{\\text{molar mass of CaO}} = \\frac{1.85 \\, \\text{g}}{56.08 \\, \\text{g\/mol}} \\approx 0.03296 \\, \\text{mol}moles of CaO=molar mass of CaOmass of CaO\u200b=56.08g\/mol1.85g\u200b\u22480.03296mol<\/p>\n\n\n\n

    Since the reaction occurs in a 1:1 molar ratio, the moles of CO2 absorbed will also be 0.03296\u2009mol0.03296 \\, \\text{mol}0.03296mol.<\/p>\n\n\n\n

    Step 3: Mass of CO2 Absorbed<\/h4>\n\n\n\n

    The molar mass of CO2 is:<\/p>\n\n\n\n

      \n
    • C=12.01\u2009g\/mol\\text{C} = 12.01 \\, \\text{g\/mol}C=12.01g\/mol<\/li>\n\n\n\n
    • O=16.00\u2009g\/mol\\text{O} = 16.00 \\, \\text{g\/mol}O=16.00g\/mol (for 2 oxygen atoms)<\/li>\n<\/ul>\n\n\n\n

      So, the molar mass of CO2 is:
      12.01+(2\u00d716.00)=44.01\u2009g\/mol12.01 + (2 \\times 16.00) = 44.01 \\, \\text{g\/mol}12.01+(2\u00d716.00)=44.01g\/mol<\/p>\n\n\n\n

      Now, the mass of CO2 absorbed is:mass of CO2=moles of CO2\u00d7molar mass of CO2=0.03296\u2009mol\u00d744.01\u2009g\/mol\u22481.45\u2009g\\text{mass of CO2} = \\text{moles of CO2} \\times \\text{molar mass of CO2} = 0.03296 \\, \\text{mol} \\times 44.01 \\, \\text{g\/mol} \\approx 1.45 \\, \\text{g}mass of CO2=moles of CO2\u00d7molar mass of CO2=0.03296mol\u00d744.01g\/mol\u22481.45g<\/p>\n\n\n\n

      Thus, 1.85 g of CaO can absorb approximately 1.45 g of CO2<\/strong>.<\/p>\n\n\n\n

      \n\n\n\n

      Volume of CO2 at STP:<\/h3>\n\n\n\n

      At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 L. Therefore, the volume of CO2 absorbed by 0.03296 mol at STP is:Volume of CO2=moles of CO2\u00d722.4\u2009L\/mol=0.03296\u2009mol\u00d722.4\u2009L\/mol\u22480.74\u2009L\\text{Volume of CO2} = \\text{moles of CO2} \\times 22.4 \\, \\text{L\/mol} = 0.03296 \\, \\text{mol} \\times 22.4 \\, \\text{L\/mol} \\approx 0.74 \\, \\text{L}Volume of CO2=moles of CO2\u00d722.4L\/mol=0.03296mol\u00d722.4L\/mol\u22480.74L<\/p>\n\n\n\n

      Thus, the CO2 would occupy approximately 0.74 L at STP<\/strong>.<\/p>\n\n\n\n

      Summary:<\/h3>\n\n\n\n
        \n
      • Mass of CO2 absorbed<\/strong>: 1.45 g<\/li>\n\n\n\n
      • Volume of CO2 at STP<\/strong>: 0.74 L<\/li>\n<\/ul>\n\n\n\n
        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        Calcium oxide can be used to “scrub” carbon dioxide from air. CaO (s) + CO2 (g) –> CaCO3 (s) What mass of CO2 could be absorbed by 1.85 g of ? Mass = g CO2 What volume would this CO2 occupy at STP? Volume = L CO2 The Correct Answer and Explanation is: Mass of […]<\/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-257713","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257713","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=257713"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257713\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257713"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257713"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257713"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}