To solve this gas stoichiometry problem, we\u2019ll follow a structured approach. The given reaction is:<\/p>\n\n\n\n
CaCO\u2083 (s) \u2192 CaO (s) + CO\u2082 (g)<\/strong><\/p>\n\n\n\n This equation is already balanced. One mole of calcium carbonate decomposes to form one mole of calcium oxide and one mole of carbon dioxide gas.<\/p>\n\n\n\n Given: 15.2 grams of CaCO\u2083Moles of CaCO\u2083=15.2 g100.09 g\/mol\u22480.152 mol\\text{Moles of CaCO\u2083} = \\frac{15.2 \\text{ g}}{100.09 \\text{ g\/mol}} \u2248 0.152 \\text{ mol}Moles of CaCO\u2083=100.09 g\/mol15.2 g\u200b\u22480.152 mol<\/p>\n\n\n\n From the balanced equation, 1 mole of CaCO\u2083 produces 1 mole of CO\u2082<\/strong>, so:Moles of CO\u2082=0.152 mol\\text{Moles of CO\u2082} = 0.152 \\text{ mol}Moles of CO\u2082=0.152 mol<\/p>\n\n\n\n At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters<\/strong>.Volume of CO\u2082=0.152 mol\u00d722.4 L\/mol=3.4048 L\\text{Volume of CO\u2082} = 0.152 \\text{ mol} \u00d7 22.4 \\text{ L\/mol} = 3.4048 \\text{ L}Volume of CO\u2082=0.152 mol\u00d722.4 L\/mol=3.4048 L<\/p>\n\n\n\n Rounded to 3 significant figures:3.40 liters of CO\u2082\\boxed{3.40 \\text{ liters of CO\u2082}}3.40 liters of CO\u2082\u200b<\/p>\n\n\n\n Gas stoichiometry involves calculations that relate quantities of reactants and products in a chemical reaction involving gases. In this problem, calcium carbonate (CaCO\u2083) decomposes when heated, producing calcium oxide (CaO) and carbon dioxide gas (CO\u2082). We are given a mass of CaCO\u2083 and asked to find the volume of CO\u2082 produced at standard temperature and pressure (STP), which is defined as 0\u00b0C and 1 atm pressure. Under these conditions, one mole of any ideal gas occupies 22.4 liters.<\/p>\n\n\n\n The first step is confirming that the equation is balanced, which it is\u2014each side of the equation contains one calcium, one carbon, and three oxygen atoms. Next, we calculate the molar mass of CaCO\u2083 by summing the atomic masses of calcium, carbon, and oxygen atoms. Using this molar mass, we convert the given mass of CaCO\u2083 into moles.<\/p>\n\n\n\n Once we have the number of moles of CaCO\u2083, we use the mole ratio from the balanced equation to determine the number of moles of CO\u2082 that will be produced. Here, the mole ratio is one to one, meaning the number of moles of CO\u2082 equals the number of moles of CaCO\u2083. Finally, we multiply the number of moles of CO\u2082 by the molar volume of a gas at STP to find the volume in liters.<\/p>\n\n\n\n The final answer is 3.40 liters of CO\u2082 gas.<\/p>\n\n\n\n Gas Stoichiometry Worksheet Laws we have learned to solve each of the following problems: Each of the chemical equations must first be balanced. Show all your work for credit: written calculation, carbon dioxide gas is released according to the following equation: CaCO3 (s) -> CaO (s) + CO2 (g) What volume of CO2 (g), measured […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-37149","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37149","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=37149"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37149\/revisions"}],"predecessor-version":[{"id":37157,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37149\/revisions\/37157"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=37149"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=37149"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=37149"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep 1: Balance the Chemical Equation<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 2: Calculate the Molar Mass of CaCO\u2083<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nStep 3: Convert Mass of CaCO\u2083 to Moles<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 4: Use Mole Ratio to Find Moles of CO\u2082<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 5: Use STP Conditions to Find Volume of CO\u2082<\/strong><\/h3>\n\n\n\n
\n\n\n\n Explanation:<\/h3>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"