The balanced chemical equation for photosynthesis is:<\/p>\n\n\n\n
6CO2+6H2O\u2192lightC6H12O6+6O26CO_2 + 6H_2O \\xrightarrow{\\text{light}} C_6H_{12}O_6 + 6O_26CO2\u200b+6H2\u200bOlight\u200bC6\u200bH12\u200bO6\u200b+6O2\u200b<\/p>\n\n\n\n
This equation tells us that 6 moles of carbon dioxide (CO\u2082) react to produce 1 mole of glucose (C\u2086H\u2081\u2082O\u2086). To determine the mass of glucose produced from 6.6g of CO\u2082, we’ll follow these steps:<\/p>\n\n\n\n
Step 1: Molar Mass of CO\u2082<\/h3>\n\n\n\n
The molar mass of CO\u2082 is calculated as:MCO2=12.01\u2009g\/mol\u2009(C)+2\u00d716.00\u2009g\/mol\u2009(O)=44.01\u2009g\/molM_{CO_2} = 12.01 \\, \\text{g\/mol} \\, (\\text{C}) + 2 \\times 16.00 \\, \\text{g\/mol} \\, (\\text{O}) = 44.01 \\, \\text{g\/mol}MCO2\u200b\u200b=12.01g\/mol(C)+2\u00d716.00g\/mol(O)=44.01g\/mol<\/p>\n\n\n\n
Step 2: Calculate Moles of CO\u2082<\/h3>\n\n\n\n
Using the mass of CO\u2082 (6.6g), we can calculate the number of moles of CO\u2082:moles of CO2=mass of CO2molar mass of CO2=6.6\u2009g44.01\u2009g\/mol=0.150\u2009mol\\text{moles of CO}_2 = \\frac{\\text{mass of CO}_2}{\\text{molar mass of CO}_2} = \\frac{6.6 \\, \\text{g}}{44.01 \\, \\text{g\/mol}} = 0.150 \\, \\text{mol}moles of CO2\u200b=molar mass of CO2\u200bmass of CO2\u200b\u200b=44.01g\/mol6.6g\u200b=0.150mol<\/p>\n\n\n\n
Step 3: Molar Mass of Glucose (C\u2086H\u2081\u2082O\u2086)<\/h3>\n\n\n\n
The molar mass of glucose (C\u2086H\u2081\u2082O\u2086) is:MC6H12O6=6\u00d712.01\u2009g\/mol+12\u00d71.008\u2009g\/mol+6\u00d716.00\u2009g\/mol=180.18\u2009g\/molM_{\\text{C}_6\\text{H}_{12}\\text{O}_6} = 6 \\times 12.01 \\, \\text{g\/mol} + 12 \\times 1.008 \\, \\text{g\/mol} + 6 \\times 16.00 \\, \\text{g\/mol} = 180.18 \\, \\text{g\/mol}MC6\u200bH12\u200bO6\u200b\u200b=6\u00d712.01g\/mol+12\u00d71.008g\/mol+6\u00d716.00g\/mol=180.18g\/mol<\/p>\n\n\n\n
Step 4: Use Mole Ratios<\/h3>\n\n\n\n
From the balanced equation, 6 moles of CO\u2082 produce 1 mole of glucose. Therefore, the moles of glucose produced from 0.150 moles of CO\u2082 will be:moles of glucose=1\u2009mol glucose6\u2009mol CO2\u00d70.150\u2009mol CO2=0.0250\u2009mol glucose\\text{moles of glucose} = \\frac{1 \\, \\text{mol glucose}}{6 \\, \\text{mol CO}_2} \\times 0.150 \\, \\text{mol CO}_2 = 0.0250 \\, \\text{mol glucose}moles of glucose=6mol CO2\u200b1mol glucose\u200b\u00d70.150mol CO2\u200b=0.0250mol glucose<\/p>\n\n\n\n
Step 5: Calculate Mass of Glucose<\/h3>\n\n\n\n
Now, we can calculate the mass of glucose produced using its molar mass:mass of glucose=moles of glucose\u00d7MC6H12O6=0.0250\u2009mol\u00d7180.18\u2009g\/mol=4.50\u2009g\\text{mass of glucose} = \\text{moles of glucose} \\times M_{\\text{C}_6\\text{H}_{12}\\text{O}_6} = 0.0250 \\, \\text{mol} \\times 180.18 \\, \\text{g\/mol} = 4.50 \\, \\text{g}mass of glucose=moles of glucose\u00d7MC6\u200bH12\u200bO6\u200b\u200b=0.0250mol\u00d7180.18g\/mol=4.50g<\/p>\n\n\n\n
Final Answer<\/h3>\n\n\n\n
The mass of glucose produced from the reaction of 6.6g of carbon dioxide is 4.5g<\/strong>.<\/p>\n\n\n\n Green plants use light from the Sun to drive photosynthesis. Photosynthesis is a chemical reaction in which water H2O and carbon dioxide CO2 chemically react to form the simple sugar glucose C6H12O6 and oxygen gas O2 . What mass of simple sugar glucose is produced by the reaction of 6.6g of carbon dioxide? Round your […]<\/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-248347","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248347","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=248347"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248347\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248347"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248347"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248347"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-209.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"