{"id":267640,"date":"2025-07-23T15:38:12","date_gmt":"2025-07-23T15:38:12","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=267640"},"modified":"2025-07-23T15:38:14","modified_gmt":"2025-07-23T15:38:14","slug":"given-the-following-reaction-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/23\/given-the-following-reaction-2\/","title":{"rendered":"Given the following reaction"},"content":{"rendered":"\n

Given the following reaction: (hint: balance the reaction first) Na2SO4+ AgNO3\u2192 + Ag2SO4 If you start with 60 g of a 3.0 M Na2SO4 solution and 40 g of a 3.0 M AgNO3 solution, a) determine the limiting reagent b) determine the number of grams of excess reagent left c)determine the number of moles of NaNO3 produced. c) determine the number of moles of Ag2SO4 produced<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Let’s break this problem down step by step:<\/p>\n\n\n\n

Step 1: Write the balanced chemical equation.<\/h3>\n\n\n\n

The unbalanced equation is:<\/p>\n\n\n\n

Na2SO4(aq)+AgNO3(aq)\u2192Ag2SO4(s)+NaNO3(aq)\\text{Na}_2\\text{SO}_4 (aq) + \\text{AgNO}_3 (aq) \\rightarrow \\text{Ag}_2\\text{SO}_4 (s) + \\text{NaNO}_3 (aq)Na2\u200bSO4\u200b(aq)+AgNO3\u200b(aq)\u2192Ag2\u200bSO4\u200b(s)+NaNO3\u200b(aq)<\/p>\n\n\n\n

Now, balance the equation:<\/p>\n\n\n\n

Na2SO4(aq)+2AgNO3(aq)\u2192Ag2SO4(s)+2NaNO3(aq)\\text{Na}_2\\text{SO}_4 (aq) + 2\\text{AgNO}_3 (aq) \\rightarrow \\text{Ag}_2\\text{SO}_4 (s) + 2\\text{NaNO}_3 (aq)Na2\u200bSO4\u200b(aq)+2AgNO3\u200b(aq)\u2192Ag2\u200bSO4\u200b(s)+2NaNO3\u200b(aq)<\/p>\n\n\n\n

Step 2: Find the moles of each reactant.<\/h3>\n\n\n\n
    \n
  • Na2SO4<\/strong>:
    Given: 60 g of Na2SO4 and a concentration of 3.0 M.
    Molar mass of Na2SO4 = 2(22.99)+32.06+4(16.00)=142.04\u2009g\/mol2(22.99) + 32.06 + 4(16.00) = 142.04 \\, \\text{g\/mol}2(22.99)+32.06+4(16.00)=142.04g\/mol.<\/li>\n<\/ul>\n\n\n\n

    To calculate the moles of Na2SO4:moles of Na2SO4=massmolar mass=60\u2009g142.04\u2009g\/mol=0.4226\u2009mol\\text{moles of Na2SO4} = \\frac{\\text{mass}}{\\text{molar mass}} = \\frac{60 \\, \\text{g}}{142.04 \\, \\text{g\/mol}} = 0.4226 \\, \\text{mol}moles of Na2SO4=molar massmass\u200b=142.04g\/mol60g\u200b=0.4226mol<\/p>\n\n\n\n

      \n
    • AgNO3<\/strong>:
      Given: 40 g of AgNO3 and a concentration of 3.0 M.
      Molar mass of AgNO3 = 107.87+14.01+3(16.00)=169.87\u2009g\/mol107.87 + 14.01 + 3(16.00) = 169.87 \\, \\text{g\/mol}107.87+14.01+3(16.00)=169.87g\/mol.<\/li>\n<\/ul>\n\n\n\n

      To calculate the moles of AgNO3:moles of AgNO3=massmolar mass=40\u2009g169.87\u2009g\/mol=0.2358\u2009mol\\text{moles of AgNO3} = \\frac{\\text{mass}}{\\text{molar mass}} = \\frac{40 \\, \\text{g}}{169.87 \\, \\text{g\/mol}} = 0.2358 \\, \\text{mol}moles of AgNO3=molar massmass\u200b=169.87g\/mol40g\u200b=0.2358mol<\/p>\n\n\n\n

      Step 3: Determine the limiting reagent.<\/h3>\n\n\n\n

      The balanced equation shows that 1 mole of Na2SO4 reacts with 2 moles of AgNO3. Therefore, for 0.4226 moles of Na2SO4, we need:0.4226\u2009mol Na2SO4\u00d72=0.8452\u2009mol AgNO30.4226 \\, \\text{mol Na2SO4} \\times 2 = 0.8452 \\, \\text{mol AgNO3}0.4226mol Na2SO4\u00d72=0.8452mol AgNO3<\/p>\n\n\n\n

      However, we only have 0.2358 moles of AgNO3, which is less than the 0.8452 moles required. Thus, AgNO3<\/strong> is the limiting reagent.<\/p>\n\n\n\n

      Step 4: Find the amount of excess reagent left (Na2SO4).<\/h3>\n\n\n\n

      From the balanced equation, 2 moles of AgNO3 react with 1 mole of Na2SO4. Since we have 0.2358 moles of AgNO3, it will react with:0.2358\u2009mol AgNO32=0.1179\u2009mol Na2SO4\\frac{0.2358 \\, \\text{mol AgNO3}}{2} = 0.1179 \\, \\text{mol Na2SO4}20.2358mol AgNO3\u200b=0.1179mol Na2SO4<\/p>\n\n\n\n

      The initial moles of Na2SO4 were 0.4226. So, the moles of Na2SO4 left over are:0.4226\u2009mol Na2SO4\u22120.1179\u2009mol Na2SO4=0.3047\u2009mol Na2SO40.4226 \\, \\text{mol Na2SO4} – 0.1179 \\, \\text{mol Na2SO4} = 0.3047 \\, \\text{mol Na2SO4}0.4226mol Na2SO4\u22120.1179mol Na2SO4=0.3047mol Na2SO4<\/p>\n\n\n\n

      Now, convert the excess moles of Na2SO4 to grams:mass of Na2SO4 left=0.3047\u2009mol\u00d7142.04\u2009g\/mol=43.2\u2009g\\text{mass of Na2SO4 left} = 0.3047 \\, \\text{mol} \\times 142.04 \\, \\text{g\/mol} = 43.2 \\, \\text{g}mass of Na2SO4 left=0.3047mol\u00d7142.04g\/mol=43.2g<\/p>\n\n\n\n

      Step 5: Find the moles of NaNO3 produced.<\/h3>\n\n\n\n

      From the balanced equation, for every 2 moles of AgNO3, 2 moles of NaNO3 are produced. Since we have 0.2358 moles of AgNO3, the moles of NaNO3 produced are:0.2358\u2009mol NaNO30.2358 \\, \\text{mol NaNO3}0.2358mol NaNO3<\/p>\n\n\n\n

      Step 6: Find the moles of Ag2SO4 produced.<\/h3>\n\n\n\n

      Again, from the balanced equation, 2 moles of AgNO3 produce 1 mole of Ag2SO4. Therefore, the moles of Ag2SO4 produced are:0.2358\u2009mol AgNO32=0.1179\u2009mol Ag2SO4\\frac{0.2358 \\, \\text{mol AgNO3}}{2} = 0.1179 \\, \\text{mol Ag2SO4}20.2358mol AgNO3\u200b=0.1179mol Ag2SO4<\/p>\n\n\n\n

      Final Answers:<\/h3>\n\n\n\n

      a) The limiting reagent is AgNO3<\/strong>.
      b) The number of grams of excess reagent (Na2SO4) left is 43.2 g<\/strong>.
      c) The number of moles of NaNO3 produced is 0.2358 mol<\/strong>.
      d) The number of moles of Ag2SO4 produced is 0.1179 mol<\/strong>.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      Given the following reaction: (hint: balance the reaction first) Na2SO4+ AgNO3\u2192 + Ag2SO4 If you start with 60 g of a 3.0 M Na2SO4 solution and 40 g of a 3.0 M AgNO3 solution, a) determine the limiting reagent b) determine the number of grams of excess reagent left c)determine the number of moles of […]<\/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-267640","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/267640","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=267640"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/267640\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=267640"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=267640"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=267640"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}