{"id":194361,"date":"2025-02-24T05:04:51","date_gmt":"2025-02-24T05:04:51","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=194361"},"modified":"2025-02-24T05:05:04","modified_gmt":"2025-02-24T05:05:04","slug":"the-overall-equation-involved-in-photosynthesis-is-co2-h2o-c6h12o6-o2-unbalanced-if-there-is-more-than-enough-water-how-many-grams-of-glucose-c6h12o6-m-180-18-g-mol-form-when-4-40-g-of-co2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/24\/the-overall-equation-involved-in-photosynthesis-is-co2-h2o-c6h12o6-o2-unbalanced-if-there-is-more-than-enough-water-how-many-grams-of-glucose-c6h12o6-m-180-18-g-mol-form-when-4-40-g-of-co2\/","title":{"rendered":"The overall equation involved in photosynthesis is CO2 + H2O C6H12O6 + O2 [unbalanced] If there is more than enough water, how many grams of glucose (C6H12O6, M = 180.18 g\/mol) form when 4.40 g of CO2 (M = 44.01 g\/mol) react"},"content":{"rendered":"\n<p>The overall equation involved in photosynthesis is CO2 + H2O C6H12O6 + O2 [unbalanced] If there is more than enough water, how many grams of glucose (C6H12O6, M = 180.18 g\/mol) form when 4.40 g of CO2 (M = 44.01 g\/mol) react? Report your answer with the correct sig figs. [30 pts.] 4) A compound is found to contain 23.3 mass % magnesium, 30.7 mass % sulfur, and 46.0 mass % oxygen. What is the empirical formula of this compound? (MMg = 24.31 g\/mol, Ms = 32.06 g\/mol, Mo = 16.00 g\/mol)<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Answers:<\/h3>\n\n\n\n<p>1) <strong>Mass of glucose (C\u2086H\u2081\u2082O\u2086) formed<\/strong> = <strong>3.00 g<\/strong> (rounded to 3 significant figures)<\/p>\n\n\n\n<p>2) <strong>Empirical formula of the compound<\/strong> = <strong>MgSO\u2083<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Photosynthesis Calculation:<\/strong><\/h4>\n\n\n\n<p>The overall balanced equation for photosynthesis is:<\/p>\n\n\n\n<p>[<br>6CO_2 + 6H_2O \\rightarrow C_6H_{12}O_6 + 6O_2<br>]<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Given <strong>4.40 g<\/strong> of CO\u2082, we first determine moles of CO\u2082:<br>[<br>\\text{Moles of CO\u2082} = \\frac{\\text{Mass of CO\u2082}}{\\text{Molar mass of CO\u2082}} = \\frac{4.40 \\text{ g}}{44.01 \\text{ g\/mol}} = 0.09998 \\text{ moles CO\u2082}<br>]<\/li>\n\n\n\n<li>From the balanced reaction, <strong>6 moles of CO\u2082 produce 1 mole of glucose<\/strong>:<br>[<br>\\text{Moles of glucose} = \\frac{0.09998}{6} = 0.016663 \\text{ moles C\u2086H\u2081\u2082O\u2086}<br>]<\/li>\n\n\n\n<li>Converting to grams:<br>[<br>\\text{Mass of glucose} = 0.016663 \\times 180.18 \\text{ g\/mol} = 3.00 \\text{ g}<br>]<\/li>\n<\/ul>\n\n\n\n<p>Thus, <strong>3.00 g of glucose<\/strong> is formed when <strong>4.40 g of CO\u2082<\/strong> reacts.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Empirical Formula Calculation:<\/strong><\/h4>\n\n\n\n<p>Given mass percentages:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Magnesium (Mg) = <strong>23.3%<\/strong><\/li>\n\n\n\n<li>Sulfur (S) = <strong>30.7%<\/strong><\/li>\n\n\n\n<li>Oxygen (O) = <strong>46.0%<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Assuming a <strong>100 g sample<\/strong>, convert mass to moles:<br>[<br>\\text{Moles of Mg} = \\frac{23.3}{24.31} = 0.958<br>]<br>[<br>\\text{Moles of S} = \\frac{30.7}{32.06} = 0.958<br>]<br>[<br>\\text{Moles of O} = \\frac{46.0}{16.00} = 2.875<br>]<\/p>\n\n\n\n<p>To find the <strong>simplest whole-number ratio<\/strong>, divide all by the smallest value (0.958):<\/p>\n\n\n\n<p>[<br>\\frac{0.958}{0.958} = 1<br>]<br>[<br>\\frac{0.958}{0.958} = 1<br>]<br>[<br>\\frac{2.875}{0.958} \\approx 3<br>]<\/p>\n\n\n\n<p>Thus, the empirical formula is <strong>MgSO\u2083<\/strong>.<\/p>\n\n\n\n<p>Now, I will generate an image representing the photosynthesis reaction.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-1304.png\" alt=\"\" class=\"wp-image-194362\"\/><\/figure>\n\n\n\n<p>I have provided an image illustrating the photosynthesis process, including the balanced equation and key elements such as carbon dioxide intake, water absorption, glucose production, and oxygen release. Let me know if you need any modifications!<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The overall equation involved in photosynthesis is CO2 + H2O C6H12O6 + O2 [unbalanced] If there is more than enough water, how many grams of glucose (C6H12O6, M = 180.18 g\/mol) form when 4.40 g of CO2 (M = 44.01 g\/mol) react? Report your answer with the correct sig figs. [30 pts.] 4) A compound [&hellip;]<\/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-194361","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194361","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=194361"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194361\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=194361"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=194361"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=194361"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}