{"id":187436,"date":"2025-02-04T18:54:20","date_gmt":"2025-02-04T18:54:20","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=187436"},"modified":"2025-02-04T18:54:22","modified_gmt":"2025-02-04T18:54:22","slug":"calculate-the-mass-of-barium-sulfate-baso4-molar-mass-233-43-g-mol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/04\/calculate-the-mass-of-barium-sulfate-baso4-molar-mass-233-43-g-mol\/","title":{"rendered":"Calculate the mass of barium sulfate (BaSO4, molar mass = 233.43 g\/mol)"},"content":{"rendered":"\n<p>Calculate the mass of barium sulfate (BaSO4, molar mass = 233.43 g\/mol) that can be produced when 350.0 mL of a 0.100-M solution of barium chloride (BaCl2) is mixed with 100.0 mL of a 0.100-M solution of iron(III) sulfate (Fe2(SO4)3)?<br>a. 2.33 g<br>b. 7.00 g<br>c. 3.50 g<br>d. 2.85 g<\/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>To determine the mass of barium sulfate (BaSO4\\text{BaSO}_4) produced, we must follow a step-by-step stoichiometric analysis of the reaction:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Write the Balanced Chemical Equation<\/strong><\/h3>\n\n\n\n<p>The reaction between barium chloride (BaCl2\\text{BaCl}_2) and iron(III) sulfate (Fe2(SO4)3\\text{Fe}_2(\\text{SO}_4)_3) is: 3BaCl2+Fe2(SO4)3\u21923BaSO4\u2193+2FeCl33\\text{BaCl}_2 + \\text{Fe}_2(\\text{SO}_4)_3 \\rightarrow 3\\text{BaSO}_4 \\downarrow + 2\\text{FeCl}_3<\/p>\n\n\n\n<p>This equation shows that <strong>3 moles of BaCl\u2082 react with 1 mole of Fe\u2082(SO\u2084)\u2083 to form 3 moles of BaSO\u2084<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Determine the Moles of Reactants<\/strong><\/h3>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Barium chloride (BaCl2\\text{BaCl}_2)<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molarity (M) = 0.100 M<\/li>\n\n\n\n<li>Volume = 350.0 mL = 0.350 L<\/li>\n\n\n\n<li>Moles of BaCl\u2082: 0.100\u00a0M\u00d70.350\u00a0L=0.0350\u00a0moles\u00a0of\u00a0BaCl20.100 \\text{ M} \\times 0.350 \\text{ L} = 0.0350 \\text{ moles of BaCl}_2<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Iron(III) sulfate (Fe2(SO4)3\\text{Fe}_2(\\text{SO}_4)_3)<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molarity (M) = 0.100 M<\/li>\n\n\n\n<li>Volume = 100.0 mL = 0.100 L<\/li>\n\n\n\n<li>Moles of Fe\u2082(SO\u2084)\u2083: 0.100\u00a0M\u00d70.100\u00a0L=0.0100\u00a0moles\u00a0of\u00a0Fe2(SO4)30.100 \\text{ M} \\times 0.100 \\text{ L} = 0.0100 \\text{ moles of Fe}_2(\\text{SO}_4)_3<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Determine the Limiting Reactant<\/strong><\/h3>\n\n\n\n<p>From the balanced equation:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>3 moles of BaCl\u2082 react with 1 mole of Fe\u2082(SO\u2084)\u2083<\/strong>.<\/li>\n\n\n\n<li>The available <strong>0.0350 moles of BaCl\u2082 would need<\/strong>: 0.0350\u00a0moles\u00a0BaCl23=0.0117\u00a0moles\u00a0Fe2(SO4)3\\frac{0.0350 \\text{ moles BaCl}_2}{3} = 0.0117 \\text{ moles Fe}_2(\\text{SO}_4)_3<\/li>\n\n\n\n<li>However, <strong>only 0.0100 moles Fe\u2082(SO\u2084)\u2083 are available<\/strong>, which is <strong>less than required<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>Thus, Fe\u2082(SO\u2084)\u2083 is the <strong>limiting reactant<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 4: Calculate the Moles of BaSO\u2084 Formed<\/strong><\/h3>\n\n\n\n<p>From the reaction, <strong>1 mole of Fe\u2082(SO\u2084)\u2083 produces 3 moles of BaSO\u2084<\/strong>: 0.0100&nbsp;moles&nbsp;Fe2(SO4)3\u00d73&nbsp;moles&nbsp;BaSO41&nbsp;mole&nbsp;Fe2(SO4)3=0.0300&nbsp;moles&nbsp;BaSO40.0100 \\text{ moles Fe}_2(\\text{SO}_4)_3 \\times \\frac{3 \\text{ moles BaSO}_4}{1 \\text{ mole Fe}_2(\\text{SO}_4)_3} = 0.0300 \\text{ moles BaSO}_4<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 5: Calculate the Mass of BaSO\u2084<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of BaSO\u2084 = <strong>233.43 g\/mol<\/strong><\/li>\n\n\n\n<li>Mass: 0.0300\u00a0moles\u00d7233.43\u00a0g\/mol=7.00\u00a0g0.0300 \\text{ moles} \\times 233.43 \\text{ g\/mol} = 7.00 \\text{ g}<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p>7.00&nbsp;g(Option\u2005\u200ab)\\mathbf{7.00 \\text{ g} \\quad (Option \\; b)}<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the mass of barium sulfate (BaSO4, molar mass = 233.43 g\/mol) that can be produced when 350.0 mL of a 0.100-M solution of barium chloride (BaCl2) is mixed with 100.0 mL of a 0.100-M solution of iron(III) sulfate (Fe2(SO4)3)?a. 2.33 gb. 7.00 gc. 3.50 gd. 2.85 g The Correct Answer and Explanation is : [&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-187436","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187436","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=187436"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187436\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187436"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187436"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187436"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}