{"id":233338,"date":"2025-06-13T04:52:19","date_gmt":"2025-06-13T04:52:19","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233338"},"modified":"2025-06-13T04:52:21","modified_gmt":"2025-06-13T04:52:21","slug":"calculations-for-benzaldehyde-acetone-and-dibenzalacetone-benzaldehyde-acetone-dibenzalacetone-molar-mass-g-mol-mass-used-or-produced-moles-g-106-12-2-650-02497-58-08-0-730-0126-234-29-1-481","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/13\/calculations-for-benzaldehyde-acetone-and-dibenzalacetone-benzaldehyde-acetone-dibenzalacetone-molar-mass-g-mol-mass-used-or-produced-moles-g-106-12-2-650-02497-58-08-0-730-0126-234-29-1-481\/","title":{"rendered":"Calculations for benzaldehyde, acetone, and dibenzalacetone benzaldehyde acetone dibenzalacetone Molar mass (g\/mol) Mass used or produced Moles (g) 106.12 2.650 .02497 58.08 0.730 .0126 234.29 1.481 .00632 (0\/1pts) From te table above, the limiting reagent is: acetone (-1 pts) Incorrect."},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Calculations for benzaldehyde, acetone, and dibenzalacetone benzaldehyde acetone dibenzalacetone Molar mass (g\/mol) Mass used or produced Moles (g) 106.12 2.650 .02497 58.08 0.730 .0126 234.29 1.481 .00632 (0\/1pts) From te table above, the limiting reagent is: acetone (-1 pts) Incorrect. Take into account both the moles of each reagent and their mole to mole ratio with the product. (0\/1pts) Theoretical yield (g) 1.450 (-1 pts) Incorrect. Make sure to determine the theoretical yield based on the limiting reagent. (0\/1pts) Percent yield (g) 25.316<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Balanced Chemical Equation:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">acetone+2\u2009benzaldehyde\u2192dibenzalacetone+2\u2009H2O\\text{acetone} + 2 \\, \\text{benzaldehyde} \\rightarrow \\text{dibenzalacetone} + 2 \\, \\text{H}_2\\text{O}acetone+2benzaldehyde\u2192dibenzalacetone+2H2\u200bO<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This shows that:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 mole of acetone reacts with 2 moles of benzaldehyde to form 1 mole of dibenzalacetone.<\/li>\n\n\n\n<li><strong>Mole ratio<\/strong>: acetone : benzaldehyde = <strong>1 : 2<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Moles of Reactants<\/h3>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Substance<\/th><th>Molar Mass (g\/mol)<\/th><th>Mass (g)<\/th><th>Moles (mol)<\/th><\/tr><\/thead><tbody><tr><td>Benzaldehyde<\/td><td>106.12<\/td><td>2.650<\/td><td>0.02497<\/td><\/tr><tr><td>Acetone<\/td><td>58.08<\/td><td>0.730<\/td><td>0.01257<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">To react with 0.01257 mol of acetone, we would need: 0.01257\u00d72=0.02514\u2009mol&nbsp;of&nbsp;benzaldehyde0.01257 \\times 2 = 0.02514 \\, \\text{mol of benzaldehyde}0.01257\u00d72=0.02514mol&nbsp;of&nbsp;benzaldehyde<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">But we only have <strong>0.02497 mol<\/strong> of benzaldehyde, which is <strong>slightly less<\/strong> than needed. Therefore:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Limiting Reagent: Benzaldehyde<\/strong><\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Theoretical Yield of Dibenzalacetone<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">From the stoichiometry:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>2 mol benzaldehyde \u2192 1 mol dibenzalacetone<\/li>\n\n\n\n<li>So, 0.02497 mol benzaldehyde \u2192<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">0.024972=0.012485\u2009mol&nbsp;dibenzalacetone&nbsp;(theoretical)\\frac{0.02497}{2} = 0.012485 \\, \\text{mol dibenzalacetone (theoretical)}20.02497\u200b=0.012485mol&nbsp;dibenzalacetone&nbsp;(theoretical)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Using molar mass of dibenzalacetone (234.29 g\/mol): Theoretical&nbsp;yield=0.012485\u00d7234.29=2.924\u2009g\\text{Theoretical yield} = 0.012485 \\times 234.29 = \\boxed{2.924 \\, \\text{g}}Theoretical&nbsp;yield=0.012485\u00d7234.29=2.924g\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Percent Yield<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Actual mass of dibenzalacetone produced = 1.481 g<br>Theoretical yield = 2.924 g Percent&nbsp;yield=(1.4812.924)\u00d7100=50.6%\\text{Percent yield} = \\left(\\frac{1.481}{2.924}\\right) \\times 100 = \\boxed{50.6\\%}Percent&nbsp;yield=(2.9241.481\u200b)\u00d7100=50.6%\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Limiting reagent<\/strong>: <strong>Benzaldehyde<\/strong><\/li>\n\n\n\n<li><strong>Theoretical yield<\/strong>: <strong>2.924 g<\/strong><\/li>\n\n\n\n<li><strong>Percent yield<\/strong>: <strong>50.6%<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Summary Explanation:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">To identify the limiting reagent in a chemical reaction, the molar amounts of each reactant must be compared using the stoichiometric ratios from the balanced equation. Although acetone had a smaller number of moles, it was not the limiting reagent because it required only half the molar amount of benzaldehyde. Since benzaldehyde was present in slightly less than the needed 2:1 ratio, it limited the amount of product that could form. From this, the theoretical yield was calculated using molar relationships, and the percent yield was found by comparing actual to theoretical yields.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-212.jpeg\" alt=\"\" class=\"wp-image-233339\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Calculations for benzaldehyde, acetone, and dibenzalacetone benzaldehyde acetone dibenzalacetone Molar mass (g\/mol) Mass used or produced Moles (g) 106.12 2.650 .02497 58.08 0.730 .0126 234.29 1.481 .00632 (0\/1pts) From te table above, the limiting reagent is: acetone (-1 pts) Incorrect. Take into account both the moles of each reagent and their mole to mole ratio [&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-233338","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233338","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=233338"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233338\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233338"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233338"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233338"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}