To solve this problem, we need to find the mass of carbon dioxide (CO2) that can be absorbed by 1.34 g of calcium oxide (CaO) based on the stoichiometry of the reaction:CaO(s)+CO2(g)\u2192CaCO3(s)\\text{CaO}(s) + \\text{CO}_2(g) \\rightarrow \\text{CaCO}_3(s)CaO(s)+CO2\u200b(g)\u2192CaCO3\u200b(s)<\/p>\n\n\n\n
Step 1: Molar Mass Calculation<\/h3>\n\n\n\n
First, we calculate the molar masses of CaO and CO2.<\/p>\n\n\n\n
- \n
- Molar mass of CaO: CaO=40.08\u2009(Ca)+16.00\u2009(O)=56.08\u2009g\/mol\\text{CaO} = 40.08 \\, (\\text{Ca}) + 16.00 \\, (\\text{O}) = 56.08 \\, \\text{g\/mol}CaO=40.08(Ca)+16.00(O)=56.08g\/mol<\/li>\n\n\n\n
- Molar mass of CO2: CO2=12.01\u2009(C)+2\u00d716.00\u2009(O)=44.01\u2009g\/mol\\text{CO}_2 = 12.01 \\, (\\text{C}) + 2 \\times 16.00 \\, (\\text{O}) = 44.01 \\, \\text{g\/mol}CO2\u200b=12.01(C)+2\u00d716.00(O)=44.01g\/mol<\/li>\n<\/ul>\n\n\n\n
Step 2: Moles of CaO<\/h3>\n\n\n\n
We can now calculate the number of moles of CaO in 1.34 g:moles of CaO=mass of CaOmolar mass of CaO=1.34\u2009g56.08\u2009g\/mol=0.0239\u2009mol\\text{moles of CaO} = \\frac{\\text{mass of CaO}}{\\text{molar mass of CaO}} = \\frac{1.34 \\, \\text{g}}{56.08 \\, \\text{g\/mol}} = 0.0239 \\, \\text{mol}moles of CaO=molar mass of CaOmass of CaO\u200b=56.08g\/mol1.34g\u200b=0.0239mol<\/p>\n\n\n\n
Step 3: Stoichiometric Ratio<\/h3>\n\n\n\n
From the balanced chemical equation, we see that the ratio of CaO to CO2 is 1:1. This means 1 mole of CaO reacts with 1 mole of CO2.<\/p>\n\n\n\n
Thus, 0.0239 moles of CaO will react with 0.0239 moles of CO2.<\/p>\n\n\n\n
Step 4: Mass of CO2<\/h3>\n\n\n\n
Now, we can calculate the mass of CO2 that will react with 0.0239 moles of CaO:mass of CO2=moles of CO2\u00d7molar mass of CO2=0.0239\u2009mol\u00d744.01\u2009g\/mol=1.051\u2009g\\text{mass of CO}_2 = \\text{moles of CO}_2 \\times \\text{molar mass of CO}_2 = 0.0239 \\, \\text{mol} \\times 44.01 \\, \\text{g\/mol} = 1.051 \\, \\text{g}mass of CO2\u200b=moles of CO2\u200b\u00d7molar mass of CO2\u200b=0.0239mol\u00d744.01g\/mol=1.051g<\/p>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
The mass of CO2 that can be absorbed by 1.34 g of CaO is 1.05 g<\/strong>.<\/p>\n\n\n\n
Explanation:<\/h3>\n\n\n\n
This calculation is based on the stoichiometric relationship between calcium oxide and carbon dioxide in the reaction. By determining the moles of CaO and using the 1:1 ratio between CaO and CO2, we can find the mass of CO2 absorbed. The key steps involve using the molar masses and applying stoichiometry to find the required mass of CO2.<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-825.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Calcium oxide can be used to “scrub” carbon dioxide from air. CaO(s) + CO2(g) \u2192 CaCO3(s) What mass of CO2 could be absorbed by 1.34 g of CaO? The Correct Answer and Explanation is: To solve this problem, we need to find the mass of carbon dioxide (CO2) that can be absorbed by 1.34 g […]<\/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-257725","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257725","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=257725"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257725\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257725"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257725"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257725"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}