{"id":200071,"date":"2025-03-13T07:20:46","date_gmt":"2025-03-13T07:20:46","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=200071"},"modified":"2025-03-13T07:20:49","modified_gmt":"2025-03-13T07:20:49","slug":"bacl2-na2so4-baso4-2nacl","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/13\/bacl2-na2so4-baso4-2nacl\/","title":{"rendered":"BaCl2 + Na2SO4 ? BaSO4 + 2NaCl"},"content":{"rendered":"\n<p>BaCl2 + Na2SO4 ? BaSO4 + 2NaCl<\/p>\n\n\n\n<p> a. How many grams of barium chloride will be required to react with 5.00 g of sodium sulfate? <\/p>\n\n\n\n<p>b. How many grams of barium chloride are required for the precipitation of 5.00 g of barium sulfate? <\/p>\n\n\n\n<p>c. How many grams of barium chloride are needed to produce 5.00 g of sodium chloride? <\/p>\n\n\n\n<p>d. How many grams of sodium sulfate are necessary for the precipitation of 5.00 g of barium chloride?<\/p>\n\n\n\n<p> e. How many grams of sodium sulfate have been added to barium chloride if 5.00 g of barium sulfate is precipitated? <\/p>\n\n\n\n<p>f. How many pounds of sodium sulfate are equivalent to 5.00 lb of sodium chloride? <\/p>\n\n\n\n<p>g. How many pounds of barium sulfate are precipitated by 5.00 lb of barium chloride? <\/p>\n\n\n\n<p>h. How many pounds of barium sulfate are precipitated by 5.00 lb of sodium sulfate? <\/p>\n\n\n\n<p>i. How many pounds of barium sulfate are equivalent to 5.00 lb of sodium chloride?<\/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>We are given the reaction:<\/p>\n\n\n\n<p>[<br>\\text{BaCl}_2 + \\text{Na}_2\\text{SO}_4 \\rightarrow \\text{BaSO}_4 + 2\\text{NaCl}<br>]<\/p>\n\n\n\n<p>This reaction is a double displacement reaction that forms barium sulfate (BaSO\u2084) as a precipitate. Let&#8217;s go through each part of the question one by one.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">a. How many grams of barium chloride will be required to react with 5.00 g of sodium sulfate?<\/h3>\n\n\n\n<p>First, calculate the molar masses of the compounds involved:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of ( \\text{Na}_2\\text{SO}_4 ) = ( 2 \\times 22.99 + 32.07 + 4 \\times 16.00 = 142.05 \\, \\text{g\/mol} )<\/li>\n\n\n\n<li>Molar mass of ( \\text{BaCl}_2 ) = ( 137.33 + 2 \\times 35.45 = 208.23 \\, \\text{g\/mol} )<\/li>\n<\/ul>\n\n\n\n<p>From the reaction, the stoichiometric ratio between sodium sulfate ((\\text{Na}_2\\text{SO}_4)) and barium chloride ((\\text{BaCl}_2)) is 1:1. Thus, the number of moles of sodium sulfate will be equal to the number of moles of barium chloride needed for the reaction.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Calculate moles of sodium sulfate:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{moles of Na}_2\\text{SO}_4 = \\frac{5.00 \\, \\text{g}}{142.05 \\, \\text{g\/mol}} = 0.0352 \\, \\text{mol}<br>]<\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li>Since the stoichiometry is 1:1, moles of barium chloride required = moles of sodium sulfate = 0.0352 mol.<\/li>\n\n\n\n<li>Convert moles of barium chloride to grams:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{grams of BaCl}_2 = 0.0352 \\, \\text{mol} \\times 208.23 \\, \\text{g\/mol} = 7.33 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>So, <strong>7.33 grams<\/strong> of barium chloride are required to react with 5.00 grams of sodium sulfate.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">b. How many grams of barium chloride are required for the precipitation of 5.00 g of barium sulfate?<\/h3>\n\n\n\n<p>Next, we use the stoichiometry of the reaction, where 1 mole of BaCl\u2082 produces 1 mole of BaSO\u2084.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Molar mass of BaSO\u2084 = ( 137.33 + 32.07 + 4 \\times 16.00 = 233.39 \\, \\text{g\/mol} ).<\/li>\n\n\n\n<li>Calculate moles of barium sulfate:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{moles of BaSO}_4 = \\frac{5.00 \\, \\text{g}}{233.39 \\, \\text{g\/mol}} = 0.0214 \\, \\text{mol}<br>]<\/p>\n\n\n\n<ol start=\"3\" class=\"wp-block-list\">\n<li>From the stoichiometry of the reaction, moles of barium chloride required = moles of barium sulfate = 0.0214 mol.<\/li>\n\n\n\n<li>Convert moles of barium chloride to grams:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{grams of BaCl}_2 = 0.0214 \\, \\text{mol} \\times 208.23 \\, \\text{g\/mol} = 4.46 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>So, <strong>4.46 grams<\/strong> of barium chloride are required to produce 5.00 grams of barium sulfate.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">c. How many grams of barium chloride are needed to produce 5.00 g of sodium chloride?<\/h3>\n\n\n\n<p>From the reaction, 1 mole of barium chloride produces 2 moles of sodium chloride. Thus, we need to calculate the number of moles of sodium chloride (NaCl) and then determine the corresponding amount of barium chloride.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Molar mass of NaCl = ( 22.99 + 35.45 = 58.44 \\, \\text{g\/mol} ).<\/li>\n\n\n\n<li>Calculate moles of sodium chloride:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{moles of NaCl} = \\frac{5.00 \\, \\text{g}}{58.44 \\, \\text{g\/mol}} = 0.0856 \\, \\text{mol}<br>]<\/p>\n\n\n\n<ol start=\"3\" class=\"wp-block-list\">\n<li>From the stoichiometry, ( 1 \\, \\text{mol} \\, \\text{BaCl}_2 ) produces 2 moles of NaCl, so moles of barium chloride required = ( \\frac{0.0856}{2} = 0.0428 \\, \\text{mol} ).<\/li>\n\n\n\n<li>Convert moles of barium chloride to grams:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{grams of BaCl}_2 = 0.0428 \\, \\text{mol} \\times 208.23 \\, \\text{g\/mol} = 8.92 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>So, <strong>8.92 grams<\/strong> of barium chloride are needed to produce 5.00 grams of sodium chloride.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">d. How many grams of sodium sulfate are necessary for the precipitation of 5.00 g of barium chloride?<\/h3>\n\n\n\n<p>Using stoichiometry, 1 mole of sodium sulfate reacts with 1 mole of barium chloride. Therefore, the number of moles of barium chloride will be equal to the number of moles of sodium sulfate.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Molar mass of BaCl\u2082 = 208.23 g\/mol.<\/li>\n\n\n\n<li>Calculate moles of barium chloride:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{moles of BaCl}_2 = \\frac{5.00 \\, \\text{g}}{208.23 \\, \\text{g\/mol}} = 0.0240 \\, \\text{mol}<br>]<\/p>\n\n\n\n<ol start=\"3\" class=\"wp-block-list\">\n<li>From the stoichiometry, moles of sodium sulfate required = moles of barium chloride = 0.0240 mol.<\/li>\n\n\n\n<li>Molar mass of sodium sulfate = 142.05 g\/mol.<\/li>\n\n\n\n<li>Convert moles of sodium sulfate to grams:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{grams of Na}_2\\text{SO}_4 = 0.0240 \\, \\text{mol} \\times 142.05 \\, \\text{g\/mol} = 3.41 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>So, <strong>3.41 grams<\/strong> of sodium sulfate are necessary for the precipitation of 5.00 grams of barium chloride.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">e. How many grams of sodium sulfate have been added to barium chloride if 5.00 g of barium sulfate is precipitated?<\/h3>\n\n\n\n<p>The stoichiometry of the reaction shows that 1 mole of sodium sulfate reacts with 1 mole of barium chloride to produce 1 mole of barium sulfate.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Molar mass of BaSO\u2084 = 233.39 g\/mol.<\/li>\n\n\n\n<li>Calculate moles of barium sulfate:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{moles of BaSO}_4 = \\frac{5.00 \\, \\text{g}}{233.39 \\, \\text{g\/mol}} = 0.0214 \\, \\text{mol}<br>]<\/p>\n\n\n\n<ol start=\"3\" class=\"wp-block-list\">\n<li>From the stoichiometry, moles of sodium sulfate required = moles of barium sulfate = 0.0214 mol.<\/li>\n\n\n\n<li>Molar mass of sodium sulfate = 142.05 g\/mol.<\/li>\n\n\n\n<li>Convert moles of sodium sulfate to grams:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{grams of Na}_2\\text{SO}_4 = 0.0214 \\, \\text{mol} \\times 142.05 \\, \\text{g\/mol} = 3.04 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>So, <strong>3.04 grams<\/strong> of sodium sulfate have been added to barium chloride if 5.00 grams of barium sulfate is precipitated.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">f. How many pounds of sodium sulfate are equivalent to 5.00 lb of sodium chloride?<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>We know that the molar ratio between sodium sulfate and sodium chloride is 1:2. First, convert pounds of sodium chloride to grams:<\/li>\n<\/ol>\n\n\n\n<p>[<br>5.00 \\, \\text{lb} = 5.00 \\times 453.592 = 2267.96 \\, \\text{g}<br>]<\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li>Calculate moles of sodium chloride:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{moles of NaCl} = \\frac{2267.96 \\, \\text{g}}{58.44 \\, \\text{g\/mol}} = 38.85 \\, \\text{mol}<br>]<\/p>\n\n\n\n<ol start=\"3\" class=\"wp-block-list\">\n<li>From the stoichiometry, moles of sodium sulfate required = ( \\frac{38.85}{2} = 19.43 \\, \\text{mol} ).<\/li>\n\n\n\n<li>Convert moles of sodium sulfate to grams:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{grams of Na}_2\\text{SO}_4 = 19.43 \\, \\text{mol} \\times 142.05 \\, \\text{g\/mol} = 2752.85 \\, \\text{g}<br>]<\/p>\n\n\n\n<ol start=\"5\" class=\"wp-block-list\">\n<li>Convert grams to pounds:<\/li>\n<\/ol>\n\n\n\n<p>[<br>2752.85 \\, \\text{g} = \\frac{2752.85}{453.592} = 6.06 \\, \\text{lb}<br>]<\/p>\n\n\n\n<p>So, <strong>6.06 pounds<\/strong> of sodium sulfate are equivalent to 5.00 lb of sodium chloride.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">g. How many pounds of barium sulfate are precipitated by 5.00 lb of barium chloride?<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Convert pounds of barium chloride to grams:<\/li>\n<\/ol>\n\n\n\n<p>[<br>5.00 \\, \\text{lb} = 5.00 \\times 453.592 = 2267.96 \\, \\text{g}<br>]<\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li>Calculate moles of barium chloride:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{moles of BaCl}_2 = \\frac{2267.96 \\, \\text{g}}{208.23 \\, \\text{g\/mol}} = 10.9 \\, \\text{mol}<br>]<\/p>\n\n\n\n<ol start=\"3\" class=\"wp-block-list\">\n<li>From the stoichiometry, moles of barium sulfate produced = moles of barium chloride = 10.9 mol.<\/li>\n\n\n\n<li>Molar mass of barium sulfate = 233.39 g\/mol.<\/li>\n\n\n\n<li>Convert moles of barium sulfate to grams:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{grams of BaSO}_4 = 10.9 \\, \\text{mol} \\times 233.39 \\, \\text{g\/mol} = 2543.9 \\, \\text{g}<br>]<\/p>\n\n\n\n<ol start=\"6\" class=\"wp-block-list\">\n<li>Convert grams to pounds:<\/li>\n<\/ol>\n\n\n\n<p>[<br>2543.9 \\, \\text{g} = \\frac{2543.9}{453.592} = 5.61 \\, \\text{lb}<br>]<\/p>\n\n\n\n<p>So, <strong>5.61 pounds<\/strong> of barium sulfate are precipitated by 5.00 lb of barium chloride.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">h. How many pounds of barium sulfate are precipitated by 5.00 lb of sodium sulfate?<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Convert pounds of sodium sulfate to grams:<\/li>\n<\/ol>\n\n\n\n<p>[<br>5.00 \\, \\text{lb} = 5.00 \\times 453.592 = 2267.96 \\, \\text{g}<br>]<\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li>Calculate moles of sodium sulfate:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{moles of Na}_2\\text{SO}_4 = \\frac{2267.96 \\, \\text{g}}{142.05 \\, \\text{g\/mol}} = 15.96 \\, \\text{mol}<br>]<\/p>\n\n\n\n<ol start=\"3\" class=\"wp-block-list\">\n<li>From the stoichiometry, moles of barium sulfate produced = moles of sodium sulfate = 15.96 mol.<\/li>\n\n\n\n<li>Molar mass of barium sulfate = 233.39 g\/mol.<\/li>\n\n\n\n<li>Convert moles of barium sulfate to grams:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{grams of BaSO}_4 = 15.96 \\, \\text{mol} \\times 233.39 \\, \\text{g\/mol} = 3726.4 \\, \\text{g}<br>]<\/p>\n\n\n\n<ol start=\"6\" class=\"wp-block-list\">\n<li>Convert grams to pounds:<\/li>\n<\/ol>\n\n\n\n<p>[<br>3726.4 \\, \\text{g} = \\frac{3726.4}{453.592} = 8.21 \\, \\text{lb}<br>]<\/p>\n\n\n\n<p>So, <strong>8.21 pounds<\/strong> of barium sulfate are precipitated by 5.00 lb of sodium sulfate.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">i. How many pounds of barium sulfate are equivalent to 5.00 lb of sodium chloride?<\/h3>\n\n\n\n<p>Using the stoichiometry, 2 moles of NaCl produce 1 mole of BaSO\u2084. Therefore, 5.00 lb of NaCl (2267.96 g) will produce half the number of moles of barium sulfate:<\/p>\n\n\n\n<p>[<br>\\text{moles of BaSO}_4 = \\frac{38.85}{2} = 19.43 \\, \\text{mol}<br>]<\/p>\n\n\n\n<p>Convert moles of BaSO\u2084 to grams and then to pounds:<\/p>\n\n\n\n<p>[<br>\\text{grams of BaSO}_4 = 19.43 \\, \\text{mol} \\times 233.39 \\, \\text{g\/mol} = 4531.1 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>[<br>4531.1 \\, \\text{g} = \\frac{4531.1}{453.592} = 10.00 \\, \\text{lb}<br>]<\/p>\n\n\n\n<p>So, <strong>10.00 pounds<\/strong> of barium sulfate are equivalent to 5.00 lb of sodium chloride.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>BaCl2 + Na2SO4 ? BaSO4 + 2NaCl a. How many grams of barium chloride will be required to react with 5.00 g of sodium sulfate? b. How many grams of barium chloride are required for the precipitation of 5.00 g of barium sulfate? c. How many grams of barium chloride are needed to produce 5.00 [&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-200071","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/200071","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=200071"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/200071\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=200071"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=200071"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=200071"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}