Give the balanced chemical equation for Cellular Respiration (all balanced chemical equations require states for full credit)<\/p>\n\n\n\n
Solid glucose (C6H12O6) & oxygen gas react to product carbon dioxide and water vapor
Find the amount of carbon dioxide produced (in g) from the conversion of 60. g glucose assuming excess oxygen gas.<\/p>\n\n\n\n
Keep track of SF
Show all steps
Calculate the amount of carbon dioxide produced (in g) given the same amount of 60. g glucose, but 60. g oxygen gas as well.Explain why your answer changed or didn’t change.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s go through this step by step.<\/p>\n\n\n\n The reaction for cellular respiration<\/strong> is:<\/p>\n\n\n\n [ Each element is balanced:<\/p>\n\n\n\n Molar mass of glucose ((\\text{C}6\\text{H}<\/em>{12}\\text{O}_6)):<\/p>\n\n\n\n [ [ From the balanced equation, 1 mole of glucose produces 6 moles of CO\u2082<\/strong>:<\/p>\n\n\n\n [ Molar mass of CO\u2082:<\/p>\n\n\n\n [ [ Final Answer (with 3 significant figures):<\/strong> Molar mass of O\u2082:<\/p>\n\n\n\n [ [ From the balanced equation, 1 mole of glucose requires 6 moles of O\u2082<\/strong>.<\/p>\n\n\n\n [ Since we only have 1.875 moles of O\u2082<\/strong>, it is the limiting reactant<\/strong>.<\/p>\n\n\n\n From the reaction, 6 moles of O\u2082 produce 6 moles of CO\u2082<\/strong>:<\/p>\n\n\n\n [ [ Final Answer (with 3 significant figures):<\/strong> In the first case, glucose was the limiting reactant<\/strong>, and oxygen was in excess. This meant that all 60.0 g of glucose was fully consumed, producing 88.0 g of CO\u2082<\/strong>.<\/p>\n\n\n\n However, in the second case, oxygen became the limiting reactant<\/strong> because there wasn’t enough of it to react with all the glucose. Instead of 2.00 moles of CO\u2082, only 1.875 moles<\/strong> of CO\u2082 could form, resulting in a lower 82.5 g of CO\u2082<\/strong>.<\/p>\n\n\n\n This demonstrates the concept of limiting reactants<\/strong> in stoichiometry\u2014when one reactant runs out first, the reaction stops, and the maximum amount of product depends on the limiting reactant. In real-world applications, ensuring sufficient reactants is essential for maximizing product yield.<\/p>\n","protected":false},"excerpt":{"rendered":" Give the balanced chemical equation for Cellular Respiration (all balanced chemical equations require states for full credit) Solid glucose (C6H12O6) & oxygen gas react to product carbon dioxide and water vaporFind the amount of carbon dioxide produced (in g) from the conversion of 60. g glucose assuming excess oxygen gas. Keep track of SFShow all […]<\/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-187338","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187338","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=187338"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187338\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187338"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187338"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187338"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Write the Balanced Chemical Equation<\/strong><\/h3>\n\n\n\n
\\text{C}6\\text{H}<\/em>{12}\\text{O}_6 (s) + 6\\text{O}_2 (g) \\rightarrow 6\\text{CO}_2 (g) + 6\\text{H}_2\\text{O} (g)
]<\/p>\n\n\n\n\n
\n\n\n\nStep 2: Find the Amount of CO\u2082 Produced from 60.0 g of Glucose (Excess O\u2082)<\/strong><\/h3>\n\n\n\n
Convert Glucose to Moles<\/strong><\/h4>\n\n\n\n
(6 \\times 12.01) + (12 \\times 1.008) + (6 \\times 16.00) = 180.16 \\, g\/mol
]<\/p>\n\n\n\n
\\frac{60.0 \\, g}{180.16 \\, g\/mol} = 0.333 \\, \\text{mol C}6\\text{H}<\/em>{12}\\text{O}_6
]<\/p>\n\n\n\nUse Stoichiometry to Find Moles of CO\u2082<\/strong><\/h4>\n\n\n\n
0.333 \\times 6 = 2.00 \\, \\text{mol CO}_2
]<\/p>\n\n\n\nConvert Moles of CO\u2082 to Grams<\/strong><\/h4>\n\n\n\n
(1 \\times 12.01) + (2 \\times 16.00) = 44.01 \\, g\/mol
]<\/p>\n\n\n\n
2.00 \\times 44.01 = 88.0 \\, g \\text{ CO}_2
]<\/p>\n\n\n\n
[
\\mathbf{88.0 \\, g \\text{ CO}_2}
]<\/p>\n\n\n\n
\n\n\n\nStep 3: Find the Amount of CO\u2082 Produced with 60.0 g Glucose and 60.0 g O\u2082<\/strong><\/h3>\n\n\n\n
Convert Oxygen to Moles<\/strong><\/h4>\n\n\n\n
2 \\times 16.00 = 32.00 \\, g\/mol
]<\/p>\n\n\n\n
\\frac{60.0 \\, g}{32.00 \\, g\/mol} = 1.875 \\, \\text{mol O}_2
]<\/p>\n\n\n\nDetermine the Limiting Reactant<\/strong><\/h4>\n\n\n\n
0.333 \\times 6 = 2.00 \\, \\text{mol O}_2 \\text{ required}
]<\/p>\n\n\n\nFind Moles of CO\u2082 from O\u2082<\/strong><\/h4>\n\n\n\n
1.875 \\times \\frac{6}{6} = 1.875 \\, \\text{mol CO}_2
]<\/p>\n\n\n\nConvert to Grams of CO\u2082<\/strong><\/h4>\n\n\n\n
1.875 \\times 44.01 = 82.5 \\, g \\text{ CO}_2
]<\/p>\n\n\n\n
[
\\mathbf{82.5 \\, g \\text{ CO}_2}
]<\/p>\n\n\n\n
\n\n\n\nStep 4: Explanation (Why Did the Answer Change?)<\/strong><\/h3>\n\n\n\n