{"id":245755,"date":"2025-07-06T12:13:19","date_gmt":"2025-07-06T12:13:19","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=245755"},"modified":"2025-07-06T12:13:22","modified_gmt":"2025-07-06T12:13:22","slug":"you-performed-experiment-following-the-balanced-chemical-equation-and-started-with-5-65-grams-of-cuso4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/06\/you-performed-experiment-following-the-balanced-chemical-equation-and-started-with-5-65-grams-of-cuso4\/","title":{"rendered":"You performed experiment following the balanced chemical equation and started with 5.65 grams of CuSO4."},"content":{"rendered":"\n
You performed experiment following the balanced chemical equation and started with 5.65 grams of CuSO4.<\/pre>\n\n\n\n
Question No. 1: 4 pt CuSO
(aq) + 2NaOH(aq)
Cu(OH)
(s) + Na
SO
(aq) (Reactants) (Products) No.1: You performed experiment following the balanced chemical equation and started with 5.65 grams of CuSO
. Calculate amount (gm) of NaOH you will need to get the maximum amount of product Cu(OH)
? Answer: Calculations: a. Mass of CuSO
(copper II sulfate) b. Formula weight of copper (II) sulfate c. Moles of CuSO
159.62 g\/mol CuSO
d. The molar ratio between CuSO
and NaOH is e. Moles of NaOH needed to reach stoichiometric quantities mol NaOH f. Formula weight of NaOH (sodium hydroxide) 40.0 g\/mol g. Mass of NaOH (sodium hydroxide) needed to reach stoichiometric quantities = g NaOH<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Let’s go through the solution step by step:<\/p>\n\n\n\n
1. Given Data<\/strong>:<\/h3>\n\n\n\n
    \n
  • Mass of CuSO\u2084 = 5.65 g<\/li>\n\n\n\n
  • Formula weight of CuSO\u2084 = 159.62 g\/mol<\/li>\n\n\n\n
  • Formula weight of NaOH = 40.0 g\/mol<\/li>\n\n\n\n
  • The balanced equation is:
    CuSO\u2084(aq) + 2NaOH(aq) \u2192 Cu(OH)\u2082(s) + Na\u2082SO\u2084(aq)<\/strong><\/li>\n<\/ul>\n\n\n\n
    2. Calculating Moles of CuSO\u2084<\/strong>:<\/h3>\n\n\n\n
    To find the moles of CuSO\u2084, we use the formula:Moles of CuSO\u2084=Mass of CuSO\u2084Molar Mass of CuSO\u2084\\text{Moles of CuSO\u2084} = \\frac{\\text{Mass of CuSO\u2084}}{\\text{Molar Mass of CuSO\u2084}}Moles of CuSO\u2084=Molar Mass of CuSO\u2084Mass of CuSO\u2084\u200bMoles of CuSO\u2084=5.65\u2009g159.62\u2009g\/mol=0.0353\u2009mol\\text{Moles of CuSO\u2084} = \\frac{5.65 \\, \\text{g}}{159.62 \\, \\text{g\/mol}} = 0.0353 \\, \\text{mol}Moles of CuSO\u2084=159.62g\/mol5.65g\u200b=0.0353mol<\/p>\n\n\n\n
    3. Molar Ratio Between CuSO\u2084 and NaOH<\/strong>:<\/h3>\n\n\n\n
    From the balanced chemical equation, the molar ratio between CuSO\u2084 and NaOH is 1:2, meaning for every 1 mole of CuSO\u2084, we need 2 moles of NaOH.Moles of NaOH required=0.0353\u2009mol CuSO\u2084\u00d72=0.0706\u2009mol NaOH\\text{Moles of NaOH required} = 0.0353 \\, \\text{mol CuSO\u2084} \\times 2 = 0.0706 \\, \\text{mol NaOH}Moles of NaOH required=0.0353mol CuSO\u2084\u00d72=0.0706mol NaOH<\/p>\n\n\n\n
    4. Calculating Mass of NaOH<\/strong>:<\/h3>\n\n\n\n
    To find the mass of NaOH required, we use the formula:Mass of NaOH=Moles of NaOH\u00d7Molar Mass of NaOH\\text{Mass of NaOH} = \\text{Moles of NaOH} \\times \\text{Molar Mass of NaOH}Mass of NaOH=Moles of NaOH\u00d7Molar Mass of NaOHMass of NaOH=0.0706\u2009mol\u00d740.0\u2009g\/mol=2.824\u2009g\\text{Mass of NaOH} = 0.0706 \\, \\text{mol} \\times 40.0 \\, \\text{g\/mol} = 2.824 \\, \\text{g}Mass of NaOH=0.0706mol\u00d740.0g\/mol=2.824g<\/p>\n\n\n\n
    5. Answer<\/strong>:<\/h3>\n\n\n\n
    You will need 2.82 grams<\/strong> of NaOH to react completely with 5.65 grams of CuSO\u2084 to form the maximum amount of Cu(OH)\u2082.<\/p>\n\n\n\n
    Explanation:<\/h3>\n\n\n\n
      \n
    1. We began with the given mass of CuSO\u2084 and converted it into moles using the molar mass.<\/li>\n\n\n\n
    2. The stoichiometric ratio between CuSO\u2084 and NaOH (from the balanced equation) told us that for every mole of CuSO\u2084, we need 2 moles of NaOH.<\/li>\n\n\n\n
    3. Using this ratio, we calculated the required moles of NaOH.<\/li>\n\n\n\n
    4. Finally, we converted the moles of NaOH into mass using the molar mass of NaOH, yielding the mass required for complete reaction.<\/li>\n<\/ol>\n\n\n\n
      This process ensures that we are using the correct stoichiometric amount of NaOH for the reaction.<\/p>\n\n\n\n
      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
      You performed experiment following the balanced chemical equation and started with 5.65 grams of CuSO4. Question No. 1: 4 pt CuSO(aq) + 2NaOH(aq)Cu(OH)(s) + NaSO(aq) (Reactants) (Products) No.1: You performed experiment following the balanced chemical equation and started with 5.65 grams of CuSO. Calculate amount (gm) of NaOH you will need to get the maximum […]<\/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-245755","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245755","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=245755"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245755\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245755"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245755"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245755"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}