{"id":192196,"date":"2025-02-17T11:16:11","date_gmt":"2025-02-17T11:16:11","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=192196"},"modified":"2025-02-17T11:16:13","modified_gmt":"2025-02-17T11:16:13","slug":"molar-mass-186-03-g-molcan-be-prepared-by-reacting-3-0-g-of-fecl2s-molar-mass-126-75-g-mol-with-5-7-g-of-cyclopentadiene-c5h6lmolar-mass-66-01-g-moland-an-excess-of-koh","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/17\/molar-mass-186-03-g-molcan-be-prepared-by-reacting-3-0-g-of-fecl2s-molar-mass-126-75-g-mol-with-5-7-g-of-cyclopentadiene-c5h6lmolar-mass-66-01-g-moland-an-excess-of-koh\/","title":{"rendered":"molar mass 186.03 g\/mol]can be prepared by reacting 3.0 g of FeCl2(s) [molar mass =126.75 g\/mol] with 5.7 g of cyclopentadiene [C5H6(l);molar mass 66.01 g\/mol]and an excess of KOH"},"content":{"rendered":"\n<p>Ferrocene, [Fe(C10H10)(s) ;molar mass 186.03 g\/mol]can be prepared by reacting 3.0 g of FeCl2(s) [molar mass =126.75 g\/mol] with 5.7 g of cyclopentadiene [C5H6(l);molar mass 66.01 g\/mol]and an excess of KOH, as shown in the following reaction \u00c3\u201a\u00c2 FeCl2\u00c3\u201a\u00c2 + 2C5H6\u00c3\u201a\u00c2 + 2KOH \u00c3\u00a2\u00e2\u20ac \u00e2\u20ac\u2122 FeC10H10\u00c3\u201a\u00c2 + 2H2O \u00c3\u201a\u00c2 A student who carried out this reaction obtained 2.7 g of ferrocene. Of the following choices which has three correct answers? \u00c3\u201a\u00c2 is limiting; 4.40 g of ferrocene could be theoretically formed; obtained 61.36 % yield<br>Ferrocene, [Fe(C10H10)(s) ;molar mass 186.03 g\/mol]can be prepared by reacting 3.0 g of FeCl2(s) [molar mass =126.75 g\/mol] with 5.7 g of cyclopentadiene [C5H6(l);molar mass 66.01 g\/mol]and an excess of KOH, as shown in the following reaction<\/p>\n\n\n\n<p>\u00c3\u201a\u00c2<\/p>\n\n\n\n<p>FeCl2\u00c3\u201a\u00c2 + 2C5H6\u00c3\u201a\u00c2 + 2KOH \u00c3\u00a2\u00e2\u20ac \u00e2\u20ac\u2122 FeC10H10\u00c3\u201a\u00c2 + 2H2O<\/p>\n\n\n\n<p>\u00c3\u201a\u00c2<\/p>\n\n\n\n<p>A student who carried out this reaction obtained 2.7 g of ferrocene. Of the following choices which has three correct answers?<\/p>\n\n\n\n<p>\u00c3\u201a\u00c2<br>is limiting; 4.40 g of ferrocene could be theoretically formed; obtained 61.36 % yield<\/p>\n\n\n\n<p>\u00c3\u201a\u00c2<br>is limiting; 2.82 g of ferrocene could be theoretically formed; obtained 95.7 % yield<\/p>\n\n\n\n<p>\u00c3\u201a\u00c2<br>is limiting; 4.2 g of ferrocene could be theoretically formed; obtained 54.6 % yield<\/p>\n\n\n\n<p>\u00c3\u201a\u00c2<br>is limiting; 0.0151 g of ferrocene could be theoretically formed; obtained 74.8 % yield<\/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<p>The correct answer is:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>FeCl2 is limiting.<\/li>\n\n\n\n<li>4.40 g of ferrocene could be theoretically formed.<\/li>\n\n\n\n<li>Obtained 61.36% yield.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Limiting Reactant<\/strong>:<br>In the given reaction, FeCl2 and cyclopentadiene (C5H6) react in a 1:2 ratio. The moles of FeCl2 are less than half the moles of C5H6, which means FeCl2 is the limiting reagent.<\/li>\n\n\n\n<li><strong>Theoretical Yield<\/strong>:<br>The limiting reagent (FeCl2) dictates the maximum amount of product that can be formed. From the moles of FeCl2 (3.0 g \/ 126.75 g\/mol = 0.0237 mol), the theoretical yield of ferrocene can be calculated using the stoichiometric ratio (1 mole of FeCl2 forms 1 mole of ferrocene).<br>The theoretical yield is therefore approximately 4.40 g of ferrocene.<\/li>\n\n\n\n<li><strong>Percent Yield<\/strong>:<br>The percent yield is calculated by dividing the actual yield (2.7 g) by the theoretical yield (4.40 g) and multiplying by 100. This results in approximately 61.36%, which matches the given answer.<\/li>\n<\/ol>\n\n\n\n<p>Now, let&#8217;s generate an image to illustrate the reaction.<\/p>\n\n\n\n<p>Here is the chemical reaction diagram for the preparation of ferrocene, showing the reactants, products, and the balanced equation. Let me know if you need any further explanations!<\/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-772.png\" alt=\"\" class=\"wp-image-192197\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Ferrocene, [Fe(C10H10)(s) ;molar mass 186.03 g\/mol]can be prepared by reacting 3.0 g of FeCl2(s) [molar mass =126.75 g\/mol] with 5.7 g of cyclopentadiene [C5H6(l);molar mass 66.01 g\/mol]and an excess of KOH, as shown in the following reaction \u00c3\u201a\u00c2 FeCl2\u00c3\u201a\u00c2 + 2C5H6\u00c3\u201a\u00c2 + 2KOH \u00c3\u00a2\u00e2\u20ac \u00e2\u20ac\u2122 FeC10H10\u00c3\u201a\u00c2 + 2H2O \u00c3\u201a\u00c2 A student who carried out this [&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-192196","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192196","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=192196"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192196\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=192196"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=192196"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=192196"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}