To solve the problems you have outlined, let’s break them down one by one.<\/p>\n\n\n\n
\n\n\n\n
Part A: Mass of Calcium Carbonate needed to produce 35.0 L of Carbon Dioxide at STP<\/h3>\n\n\n\n
The decomposition reaction of calcium carbonate (CaCO\u2083) is:<\/p>\n\n\n\n
[ \\text{CaCO\u2083 (s)} \\rightarrow \\text{CaO (s)} + \\text{CO\u2082 (g)} ]<\/p>\n\n\n\n
To calculate the mass of calcium carbonate needed to produce 35.0 L of CO\u2082 at STP, we will follow these steps:<\/p>\n\n\n\n
- \n
- Use the ideal gas law<\/strong> to find moles of CO\u2082 produced:
[
PV = nRT
]
At STP (Standard Temperature and Pressure), the molar volume of a gas is 22.4 L. Therefore, the moles of CO\u2082 can be calculated as:
[
n_{\\text{CO\u2082}} = \\frac{V_{\\text{CO\u2082}}}{22.4 \\, \\text{L\/mol}} = \\frac{35.0 \\, \\text{L}}{22.4 \\, \\text{L\/mol}} = 1.56 \\, \\text{mol CO\u2082}
]<\/li>\n\n\n\n- Use stoichiometry<\/strong> to determine moles of CaCO\u2083:
The balanced reaction shows that 1 mole of CaCO\u2083 produces 1 mole of CO\u2082. Therefore:
[
n_{\\text{CaCO\u2083}} = 1.56 \\, \\text{mol CaCO\u2083}
]<\/li>\n\n\n\n- Calculate the mass of CaCO\u2083<\/strong>:
The molar mass of CaCO\u2083 is approximately 100.1 g\/mol. So, the mass of CaCO\u2083 required is:
[
\\text{Mass of CaCO\u2083} = n_{\\text{CaCO\u2083}} \\times \\text{Molar mass of CaCO\u2083}
]
[
\\text{Mass of CaCO\u2083} = 1.56 \\, \\text{mol} \\times 100.1 \\, \\text{g\/mol} = 156.2 \\, \\text{g}
]<\/li>\n<\/ol>\n\n\n\nThus, 156.2 g of calcium carbonate<\/strong> is needed to produce 35.0 L of CO\u2082 at STP.<\/p>\n\n\n\n
\n\n\n\nPart B: Volume of CO\u2082 produced by combustion of 2.20 g of Butane<\/h3>\n\n\n\n
The balanced equation for the combustion of butane (C\u2084H\u2081\u2080) is:<\/p>\n\n\n\n
[ 2 \\, \\text{C\u2084H\u2081\u2080 (g)} + 13 \\, \\text{O\u2082 (g)} \\rightarrow 8 \\, \\text{CO\u2082 (g)} + 10 \\, \\text{H\u2082O (l)} ]<\/p>\n\n\n\n
- \n
- Calculate the moles of butane (C\u2084H\u2081\u2080)<\/strong>:
The molar mass of butane is approximately 58.12 g\/mol. Using the given mass of butane:
[
n_{\\text{C\u2084H\u2081\u2080}} = \\frac{\\text{Mass of C\u2084H\u2081\u2080}}{\\text{Molar mass of C\u2084H\u2081\u2080}} = \\frac{2.20 \\, \\text{g}}{58.12 \\, \\text{g\/mol}} = 0.0379 \\, \\text{mol C\u2084H\u2081\u2080}
]<\/li>\n\n\n\n- Use stoichiometry to find the moles of CO\u2082<\/strong>:
From the balanced equation, we know that 2 moles of C\u2084H\u2081\u2080 produce 8 moles of CO\u2082. Therefore:
[
n_{\\text{CO\u2082}} = \\frac{8}{2} \\times n_{\\text{C\u2084H\u2081\u2080}} = 4 \\times 0.0379 \\, \\text{mol} = 0.1516 \\, \\text{mol CO\u2082}
]<\/li>\n\n\n\n- Use the ideal gas law<\/strong> to calculate the volume of CO\u2082 at the given conditions:
At 1.00 atm and 23\u00b0C (which is 296 K), the molar volume of an ideal gas can be calculated using the ideal gas law:
[
V = nRT \/ P
]
Where:<\/li>\n<\/ol>\n\n\n\n- \n
- (n = 0.1516 \\, \\text{mol CO\u2082})<\/li>\n\n\n\n
- (R = 0.0821 \\, \\text{L\u00b7atm\/(mol\u00b7K)})<\/li>\n\n\n\n
- (T = 296 \\, \\text{K})<\/li>\n\n\n\n
- (P = 1.00 \\, \\text{atm}) Substituting the values:
[
V = \\frac{(0.1516 \\, \\text{mol}) \\times (0.0821 \\, \\text{L\u00b7atm\/mol\u00b7K}) \\times (296 \\, \\text{K})}{1.00 \\, \\text{atm}} = 3.69 \\, \\text{L}
]<\/li>\n<\/ul>\n\n\n\nThus, 3.69 L of CO\u2082<\/strong> is produced by the combustion of 2.20 g of butane at 1.00 atm and 23\u00b0C.<\/p>\n\n\n\n
\n\n\n\nVisual Representation<\/h3>\n\n\n\n
Now, I will generate an image representing the steps involved in these calculations.<\/p>\n\n\n\n
Here is the visual representation of the calculations for both problems. The diagram outlines the chemical reactions, stoichiometry, and conversion factors for each, along with the steps to calculate the mass of calcium carbonate and the volume of carbon dioxide produced. Let me know if you need further clarification on any part of the solution!<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-794.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction CaCO3(s)\u00c3\u00a2\u00e2\u20ac \u00e2\u20ac\u2122CaO(s) + CO\u00c3\u00a2\u00e2\u20ac\u0161\u00e2\u20ac\u0161(g) What is the mass of calcium carbonate needed to produce 35.0 L of carbon dioxide at STP? Express your answer with the appropriate units. View Available Hint(s) 0 mass of CaCO3 = 370.8 Part B […]<\/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-192269","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192269","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=192269"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192269\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=192269"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=192269"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=192269"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Use stoichiometry to find the moles of CO\u2082<\/strong>:
- Use stoichiometry<\/strong> to determine moles of CaCO\u2083: