{"id":239322,"date":"2025-07-02T19:38:13","date_gmt":"2025-07-02T19:38:13","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=239322"},"modified":"2025-07-02T19:38:16","modified_gmt":"2025-07-02T19:38:16","slug":"what-mass-of-barium-sulfate-molar-mass-233-g-mol-is-produced-when-125-ml-of-a-0-150-m-solution-of-barium-chloride-is-mixed-with-125-ml-of-a-0-150-m-solution-of-ironiii-sulfate","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/02\/what-mass-of-barium-sulfate-molar-mass-233-g-mol-is-produced-when-125-ml-of-a-0-150-m-solution-of-barium-chloride-is-mixed-with-125-ml-of-a-0-150-m-solution-of-ironiii-sulfate\/","title":{"rendered":"What mass of barium sulfate (molar mass = 233 g\/mol) is produced when 125 mL of a 0.150 M solution of barium chloride is mixed with 125 mL of a 0.150 M solution of iron(III) sulfate"},"content":{"rendered":"\n<p>What mass of barium sulfate (molar mass = 233 g\/mol) is produced when 125 mL of a 0.150 M solution of barium chloride is mixed with 125 mL of a 0.150 M solution of iron(III) sulfate?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>The chemical reaction between barium chloride (BaCl\u2082) and iron(III) sulfate (Fe\u2082(SO\u2084)\u2083) forms barium sulfate (BaSO\u2084) and iron(III) chloride (FeCl\u2083) as products. The balanced equation for this reaction is: 3\u2009BaCl2(aq)+Fe2(SO4)3(aq)\u21923\u2009BaSO4(s)+2\u2009FeCl3(aq)3 \\, \\text{BaCl}_2 (aq) + Fe_2(SO_4)_3 (aq) \\rightarrow 3 \\, \\text{BaSO}_4 (s) + 2 \\, \\text{FeCl}_3 (aq)<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Moles of barium chloride and iron(III) sulfate<\/h3>\n\n\n\n<p>We are given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of barium chloride solution = 125 mL = 0.125 L<\/li>\n\n\n\n<li>Concentration of barium chloride solution = 0.150 M<\/li>\n\n\n\n<li>Volume of iron(III) sulfate solution = 125 mL = 0.125 L<\/li>\n\n\n\n<li>Concentration of iron(III) sulfate solution = 0.150 M<\/li>\n<\/ul>\n\n\n\n<p>To find the moles of each reactant, we use the formula: Moles=Molarity\u00d7Volume&nbsp;(L)\\text{Moles} = \\text{Molarity} \\times \\text{Volume (L)}<\/p>\n\n\n\n<p>For barium chloride: Moles&nbsp;of&nbsp;BaCl2=0.150\u2009M\u00d70.125\u2009L=0.01875\u2009moles\\text{Moles of BaCl}_2 = 0.150 \\, \\text{M} \\times 0.125 \\, \\text{L} = 0.01875 \\, \\text{moles}<\/p>\n\n\n\n<p>For iron(III) sulfate: Moles&nbsp;of&nbsp;Fe2(SO4)3=0.150\u2009M\u00d70.125\u2009L=0.01875\u2009moles\\text{Moles of Fe}_2(SO_4)_3 = 0.150 \\, \\text{M} \\times 0.125 \\, \\text{L} = 0.01875 \\, \\text{moles}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Limiting reactant<\/h3>\n\n\n\n<p>From the balanced equation, the stoichiometric ratio between barium chloride and iron(III) sulfate is 3:1. For every 3 moles of BaCl\u2082, 1 mole of Fe\u2082(SO\u2084)\u2083 is required. Since both reactants are present in equal amounts (0.01875 moles), iron(III) sulfate will be the limiting reactant because only 1 mole of Fe\u2082(SO\u2084)\u2083 is required for every 3 moles of BaCl\u2082.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Moles of barium sulfate produced<\/h3>\n\n\n\n<p>Since the limiting reactant is iron(III) sulfate, the number of moles of barium sulfate produced will be determined by the moles of Fe\u2082(SO\u2084)\u2083. From the balanced equation, 1 mole of Fe\u2082(SO\u2084)\u2083 produces 3 moles of BaSO\u2084. Moles&nbsp;of&nbsp;BaSO4=3\u00d7Moles&nbsp;of&nbsp;Fe2(SO4)3=3\u00d70.01875=0.05625\u2009moles\\text{Moles of BaSO}_4 = 3 \\times \\text{Moles of Fe}_2(SO_4)_3 = 3 \\times 0.01875 = 0.05625 \\, \\text{moles}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Mass of barium sulfate<\/h3>\n\n\n\n<p>To calculate the mass of barium sulfate, we multiply the number of moles of BaSO\u2084 by its molar mass: Mass&nbsp;of&nbsp;BaSO4=Moles&nbsp;of&nbsp;BaSO4\u00d7Molar&nbsp;Mass&nbsp;of&nbsp;BaSO4\\text{Mass of BaSO}_4 = \\text{Moles of BaSO}_4 \\times \\text{Molar Mass of BaSO}_4 Mass&nbsp;of&nbsp;BaSO4=0.05625\u2009moles\u00d7233\u2009g\/mol=13.12\u2009g\\text{Mass of BaSO}_4 = 0.05625 \\, \\text{moles} \\times 233 \\, \\text{g\/mol} = 13.12 \\, \\text{g}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The mass of barium sulfate produced is <strong>13.12 g<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>What mass of barium sulfate (molar mass = 233 g\/mol) is produced when 125 mL of a 0.150 M solution of barium chloride is mixed with 125 mL of a 0.150 M solution of iron(III) sulfate? The correct answer and explanation is: The chemical reaction between barium chloride (BaCl\u2082) and iron(III) sulfate (Fe\u2082(SO\u2084)\u2083) forms barium [&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-239322","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239322","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=239322"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239322\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=239322"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=239322"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=239322"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}