{"id":217913,"date":"2025-05-22T17:09:05","date_gmt":"2025-05-22T17:09:05","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=217913"},"modified":"2025-05-22T19:23:15","modified_gmt":"2025-05-22T19:23:15","slug":"please-answer-all-questions-showing-work-aa","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/22\/please-answer-all-questions-showing-work-aa\/","title":{"rendered":"What volume of 0.175 M Na3PO4 solution is required to completely react with 126mL of 0.150 M 0.102 M CuCl2"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Solution Stoichiometry: Please answer all questions, showing work.\u00c3\u201a\u00c2<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A. What volume of 0.175 M Na3PO4 solution is required to completely react with 126mL of 0.150 M 0.102 M CuCl2?<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">2 Na3PO4 (aq) + 3 CuCl2 (aq) -&gt; Cu3(PO4)2 (s) + 6 NaCl (aq)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">B. What is the Molarity of ZnCl2 that forms when 25.0 g of zinc completely react with a solution of CuCl2? Assume a final volume of 265 mL.\u00c3\u201a\u00c2<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">2 Al (s) + 3 H2SO4 (aq) -&gt; Al2(SO4)3 (aq) + 3 H2 (g)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">C. What volume (in mL) of a 0.150 M HNO3 solution will completely react with 35.7 mL of a 0.108 M sodium carbonate solution?\u00c3\u201a\u00c2<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Na2CO3 (aq) + 2HNO3 (aq) -&gt; 2 NaNO3 (aq) + CO2 (g) + H2O (l)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">What mass of carbon dioxide forms?\u00c3\u201a\u00c2<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">A. Volume of 0.175 M Na\u2083PO\u2084 needed to react with 126 mL of 0.150 M CuCl\u2082<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Reaction:<br>$2 \\text{Na}_3\\text{PO}_4 (aq) + 3 \\text{CuCl}_2 (aq) \\to \\text{Cu}_3(\\text{PO}_4)_2 (s) + 6 \\text{NaCl} (aq)$<\/li>\n\n\n\n<li>Volume of CuCl\u2082 = 126 mL = 0.126 L<\/li>\n\n\n\n<li>Molarity of CuCl\u2082 = 0.150 M<\/li>\n\n\n\n<li>Molarity of Na\u2083PO\u2084 = 0.175 M<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Step 1:<\/strong> Calculate moles of CuCl\u2082<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{moles CuCl}_2 = M \\times V = 0.150 \\, \\text{mol\/L} \\times 0.126 \\, \\text{L} = 0.0189 \\, \\text{mol}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Step 2:<\/strong> Use mole ratio from balanced equation to find moles of Na\u2083PO\u2084 needed:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\frac{2 \\, \\text{mol Na}_3\\text{PO}_4}{3 \\, \\text{mol CuCl}_2} = x \\implies x = 0.0189 \\times \\frac{2}{3} = 0.0126 \\, \\text{mol Na}_3\\text{PO}_4<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Step 3:<\/strong> Calculate volume of 0.175 M Na\u2083PO\u2084 solution that contains 0.0126 mol:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>V = \\frac{\\text{moles}}{M} = \\frac{0.0126}{0.175} = 0.072 \\, \\text{L} = 72.0 \\, \\text{mL}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Answer A:<\/strong> 72.0 mL of 0.175 M Na\u2083PO\u2084 solution is required.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">B. Molarity of ZnCl\u2082 formed from 25.0 g zinc reacting with CuCl\u2082 solution, volume = 265 mL<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Note: The reaction for zinc with CuCl\u2082 was not provided, but typically zinc displaces copper in:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{Zn (s)} + \\text{CuCl}_2 (aq) \\to \\text{ZnCl}_2 (aq) + \\text{Cu (s)}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass Zn = 25.0 g<\/li>\n\n\n\n<li>Volume final solution = 265 mL = 0.265 L<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Step 1:<\/strong> Calculate moles of zinc:<br>Atomic mass Zn = 65.38 g\/mol<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{moles Zn} = \\frac{25.0}{65.38} = 0.3823 \\, \\text{mol}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Step 2:<\/strong> From the reaction, 1 mole Zn produces 1 mole ZnCl\u2082, so:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{moles ZnCl}_2 = 0.3823 \\, \\text{mol}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Step 3:<\/strong> Calculate molarity of ZnCl\u2082:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>M = \\frac{\\text{moles}}{\\text{volume in liters}} = \\frac{0.3823}{0.265} = 1.44 \\, \\text{M}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Answer B:<\/strong> The molarity of ZnCl\u2082 formed is 1.44 M.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">C. Volume of 0.150 M HNO\u2083 required to react with 35.7 mL of 0.108 M Na\u2082CO\u2083, and mass of CO\u2082 formed<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Reaction:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{Na}_2\\text{CO}_3 (aq) + 2 \\text{HNO}_3 (aq) \\to 2 \\text{NaNO}_3 (aq) + \\text{CO}_2 (g) + \\text{H}_2\\text{O} (l)<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume Na\u2082CO\u2083 = 35.7 mL = 0.0357 L<\/li>\n\n\n\n<li>Molarity Na\u2082CO\u2083 = 0.108 M<\/li>\n\n\n\n<li>Molarity HNO\u2083 = 0.150 M<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Step 1:<\/strong> Calculate moles Na\u2082CO\u2083:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{moles Na}_2\\text{CO}_3 = 0.108 \\times 0.0357 = 0.003856 \\, \\text{mol}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Step 2:<\/strong> Use mole ratio for HNO\u2083 to Na\u2082CO\u2083 (2:1):<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{moles HNO}_3 = 0.003856 \\times 2 = 0.007712 \\, \\text{mol}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Step 3:<\/strong> Calculate volume of 0.150 M HNO\u2083 needed:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>V = \\frac{\\text{moles}}{M} = \\frac{0.007712}{0.150} = 0.05141 \\, \\text{L} = 51.4 \\, \\text{mL}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Mass of CO\u2082 formed<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">From reaction, 1 mole Na\u2082CO\u2083 produces 1 mole CO\u2082.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Step 4:<\/strong> Moles CO\u2082 = moles Na\u2082CO\u2083 = 0.003856 mol<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Step 5:<\/strong> Calculate mass CO\u2082:<br>Molar mass CO\u2082 = 44.01 g\/mol<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{mass CO}_2 = 0.003856 \\times 44.01 = 0.1697 \\, \\text{g}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Answer C:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of HNO\u2083 required = 51.4 mL<\/li>\n\n\n\n<li>Mass of CO\u2082 formed = 0.170 g<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Part<\/th><th>Answer<\/th><\/tr><\/thead><tbody><tr><td>A<\/td><td>72.0 mL of 0.175 M Na\u2083PO\u2084<\/td><\/tr><tr><td>B<\/td><td>1.44 M ZnCl\u2082 solution formed<\/td><\/tr><tr><td>C<\/td><td>51.4 mL of 0.150 M HNO\u2083; 0.170 g CO\u2082 formed<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Stoichiometry allows us to calculate the amounts of reactants and products in chemical reactions by using balanced chemical equations. Each equation provides the mole ratios between substances, which is key to solving these problems.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>In part A<\/strong>, the balanced equation $2 \\text{Na}_3\\text{PO}_4 + 3 \\text{CuCl}_2 \\to \\text{Cu}_3(\\text{PO}_4)_2 + 6 \\text{NaCl}$ tells us that 2 moles of sodium phosphate react with 3 moles of copper chloride. Given the molarity and volume of CuCl\u2082, we find moles of CuCl\u2082 and then calculate the corresponding moles of Na\u2083PO\u2084 using the mole ratio. Finally, converting moles of Na\u2083PO\u2084 to volume using its molarity yields the volume required.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Part B<\/strong> involves a displacement reaction where zinc metal reacts with CuCl\u2082 solution forming ZnCl\u2082 and copper metal. Since 1 mole of zinc produces 1 mole of ZnCl\u2082, the moles of zinc directly give moles of ZnCl\u2082. Using the final solution volume, we compute molarity as moles\/volume. This approach applies conservation of mass and mole equivalency in reactions.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Part C<\/strong> requires reacting sodium carbonate with nitric acid. The balanced equation shows that 2 moles of HNO\u2083 react with 1 mole of Na\u2082CO\u2083. Calculating moles of Na\u2082CO\u2083 from given concentration and volume, then doubling it to find moles of HNO\u2083 needed, allows us to find the volume of HNO\u2083 required. Additionally, 1 mole of Na\u2082CO\u2083 produces 1 mole of CO\u2082 gas. Converting moles of CO\u2082 to grams using molar mass gives the mass of carbon dioxide formed.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Understanding mole ratios, molarity, and volume relationships is critical in stoichiometry. These calculations are foundational in chemistry, especially for preparing solutions, reacting chemicals quantitatively, and predicting yields.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-104.png\" alt=\"\" class=\"wp-image-217914\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Solution Stoichiometry: Please answer all questions, showing work.\u00c3\u201a\u00c2 A. What volume of 0.175 M Na3PO4 solution is required to completely react with 126mL of 0.150 M 0.102 M CuCl2? 2 Na3PO4 (aq) + 3 CuCl2 (aq) -&gt; Cu3(PO4)2 (s) + 6 NaCl (aq) B. What is the Molarity of ZnCl2 that forms when 25.0 g [&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-217913","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217913","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=217913"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217913\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217913"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217913"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217913"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}