I. Write the chemical reaction involving the synthesis of 4-methylcyclohexene and Calculate their molar mass. Density of 4-methylcyclohexanol Density of phosphoric acid 1.69 g\/mL 0.916 g\/mL In an empty 5 mL conical vial, 2.7 mL of the 4-methylcyclohexanol and 2.3 mL of phosphoric acid are mixed. (DM\/V) (a) How many grams of each reagent is present? (Show your work) Jenny X (b) How many moles of each reagent is present? (Show your work)<\/p>\n\n\n\n
(c) Which reagent is the Limiting Reagent? Why? (Show your work) (d) What is the theoretical yield of the “4-methylcyclohexene”? (Show your work) (e) A student reports a 62.5% yield. How much product did the student make? (Show your work)<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To tackle this problem, let’s break it down step by step. The chemical reaction for synthesizing 4-methylcyclohexene involves the dehydration of 4-methylcyclohexanol in the presence of phosphoric acid. This is an example of an elimination reaction.<\/p>\n\n\n\n The dehydration reaction for 4-methylcyclohexene is as follows: 4-methylcyclohexanol\u2192H3PO44-methylcyclohexene+H2O\\text{4-methylcyclohexanol} \\xrightarrow{\\text{H}_3\\text{PO}_4} \\text{4-methylcyclohexene} + \\text{H}_2\\text{O}<\/p>\n\n\n\n To calculate the molar mass of 4-methylcyclohexene, we sum up the atomic masses of its components:<\/p>\n\n\n\n The molecular formula for 4-methylcyclohexene is C7H12C_7H_{12}. So the molar mass of 4-methylcyclohexene is: 7(12.01\u2009g\/mol)+12(1.008\u2009g\/mol)=84.07\u2009g\/mol+12.096\u2009g\/mol=96.166\u2009g\/mol7(12.01 \\, \\text{g\/mol}) + 12(1.008 \\, \\text{g\/mol}) = 84.07 \\, \\text{g\/mol} + 12.096 \\, \\text{g\/mol} = 96.166 \\, \\text{g\/mol}<\/p>\n\n\n\n Given that:<\/p>\n\n\n\n Now calculate the mass of each reagent using the formula Mass=Density\u00d7Volume\\text{Mass} = \\text{Density} \\times \\text{Volume}.<\/p>\n\n\n\n For 4-methylcyclohexanol<\/strong>: Mass of 4-methylcyclohexanol=1.69\u2009g\/mL\u00d72.7\u2009mL=4.563\u2009g\\text{Mass of 4-methylcyclohexanol} = 1.69 \\, \\text{g\/mL} \\times 2.7 \\, \\text{mL} = 4.563 \\, \\text{g}<\/p>\n\n\n\n For phosphoric acid<\/strong>: Mass of phosphoric acid=0.916\u2009g\/mL\u00d72.3\u2009mL=2.107\u2009g\\text{Mass of phosphoric acid} = 0.916 \\, \\text{g\/mL} \\times 2.3 \\, \\text{mL} = 2.107 \\, \\text{g}<\/p>\n\n\n\n To find the number of moles, use the formula Moles=MassMolar Mass\\text{Moles} = \\frac{\\text{Mass}}{\\text{Molar Mass}}.<\/p>\n\n\n\n Molar Mass of 4-methylcyclohexanol<\/strong>: The molar mass of 4-methylcyclohexanol is 114.18 g\/mol (C7H14O).<\/p>\n\n\n\n Moles of 4-methylcyclohexanol<\/strong>: Moles of 4-methylcyclohexanol=4.563\u2009g114.18\u2009g\/mol=0.0399\u2009mol\\text{Moles of 4-methylcyclohexanol} = \\frac{4.563 \\, \\text{g}}{114.18 \\, \\text{g\/mol}} = 0.0399 \\, \\text{mol}<\/p>\n\n\n\n Molar Mass of Phosphoric Acid (H\u2083PO\u2084)<\/strong>: The molar mass of phosphoric acid is 98.00 g\/mol.<\/p>\n\n\n\n Moles of phosphoric acid<\/strong>: Moles of phosphoric acid=2.107\u2009g98.00\u2009g\/mol=0.0215\u2009mol\\text{Moles of phosphoric acid} = \\frac{2.107 \\, \\text{g}}{98.00 \\, \\text{g\/mol}} = 0.0215 \\, \\text{mol}<\/p>\n\n\n\n To determine the limiting reagent, we need to look at the stoichiometry of the reaction. The balanced equation shows that 1 mole of 4-methylcyclohexanol reacts with 1 mole of phosphoric acid.<\/p>\n\n\n\n We have:<\/p>\n\n\n\n Since phosphoric acid is present in a smaller amount, it is the limiting reagent.<\/p>\n\n\n\n Since the limiting reagent is phosphoric acid, it will determine the maximum number of moles of 4-methylcyclohexene that can be formed. According to the stoichiometry, 1 mole of phosphoric acid produces 1 mole of 4-methylcyclohexene.<\/p>\n\n\n\n Thus, the theoretical yield in moles of 4-methylcyclohexene is 0.0215 mol.<\/p>\n\n\n\n To find the theoretical yield in grams: Mass of 4-methylcyclohexene=0.0215\u2009mol\u00d796.166\u2009g\/mol=2.0676\u2009g\\text{Mass of 4-methylcyclohexene} = 0.0215 \\, \\text{mol} \\times 96.166 \\, \\text{g\/mol} = 2.0676 \\, \\text{g}<\/p>\n\n\n\n The student reports a 62.5% yield, meaning that the actual yield is 62.5% of the theoretical yield: Actual Yield=62.5%\u00d72.0676\u2009g=1.292\u2009g\\text{Actual Yield} = 62.5\\% \\times 2.0676 \\, \\text{g} = 1.292 \\, \\text{g}<\/p>\n\n\n\n I. Write the chemical reaction involving the synthesis of 4-methylcyclohexene and Calculate their molar mass. Density of 4-methylcyclohexanol Density of phosphoric acid 1.69 g\/mL 0.916 g\/mL In an empty 5 mL conical vial, 2.7 mL of the 4-methylcyclohexanol and 2.3 mL of phosphoric acid are mixed. (DM\/V) (a) How many grams of each reagent is […]<\/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-186044","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186044","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=186044"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186044\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186044"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186044"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186044"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}1. Chemical reaction:<\/h3>\n\n\n\n
2. Molar Mass of 4-methylcyclohexene:<\/h3>\n\n\n\n
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3. Mass Calculation for Reagents:<\/h3>\n\n\n\n
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4. Moles of Reagents:<\/h3>\n\n\n\n
5. Limiting Reagent:<\/h3>\n\n\n\n
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6. Theoretical Yield of 4-Methylcyclohexene:<\/h3>\n\n\n\n
7. Actual Yield:<\/h3>\n\n\n\n
Conclusion:<\/h3>\n\n\n\n
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