{"id":183871,"date":"2025-01-17T07:32:43","date_gmt":"2025-01-17T07:32:43","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183871"},"modified":"2025-01-17T07:32:45","modified_gmt":"2025-01-17T07:32:45","slug":"for-the-reaction-of-hydrochloric-acid-and-calcium-carbonate","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/17\/for-the-reaction-of-hydrochloric-acid-and-calcium-carbonate\/","title":{"rendered":"For the reaction of hydrochloric acid and calcium carbonate"},"content":{"rendered":"\n

For the reaction of hydrochloric acid and calcium carbonate, the balanced equation is 2 HCl + CaCO3 \u00c3\u00a2\u00e2\u20ac \u00e2\u20ac\u2122 CO2 + H2O + CaCl2 If 0.400 mol HCl is reacted with 0.300 mol CaCO3: a. Which reactant is limiting?\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9___<\/em><\/strong><\/em><\/strong> b. How many moles of CO2 can be formed?\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9___<\/em><\/strong><\/em><\/strong> c. If 0.180 mol of CO2 are actually formed, what is the percent yield? \u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9 For the reaction of hydrochloric acid and calcium carbonate, the balanced equation is 2 HCl + CaCO3 \u00c3\u00a2\u00e2\u20ac \u00e2\u20ac\u2122 CO2 + H2O + CaCl2 If 0.400 mol HCl is reacted with 0.300 mol CaCO3: a. Which reactant is limiting?\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9___<\/em><\/strong><\/em><\/strong> b. How many moles of CO2 can be formed?\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9___<\/em><\/strong><\/em><\/strong> c. If 0.180 mol of CO2 are actually formed, what is the percent yield? \u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9____<\/em><\/strong><\/em><\/strong><\/p>\n\n\n\n

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

Reaction:<\/h3>\n\n\n\n

[ 2 \\text{HCl} + \\text{CaCO}_3 \\rightarrow \\text{CO}_2 + \\text{H}_2\\text{O} + \\text{CaCl}_2 ]<\/p>\n\n\n\n

Step-by-step solution:<\/h4>\n\n\n\n

(a) Identify the limiting reactant<\/strong><\/h4>\n\n\n\n
    \n
  • Mole ratio from the balanced equation<\/strong>:
    2 mol HCl reacts with 1 mol CaCO\u2083.<\/li>\n\n\n\n
  • Moles of HCl available<\/strong>: 0.400 mol<\/li>\n\n\n\n
  • Moles of CaCO\u2083 available<\/strong>: 0.300 mol<\/li>\n\n\n\n
  • Calculate how much HCl is needed to react with 0.300 mol CaCO\u2083<\/strong>:
    [
    \\text{Moles of HCl required} = 0.300 \\, \\text{mol} \\times \\frac{2 \\, \\text{mol HCl}}{1 \\, \\text{mol CaCO}_3} = 0.600 \\, \\text{mol HCl}
    ]
    Since only 0.400 mol of HCl is available (less than 0.600 mol), HCl is the limiting reactant<\/strong>.<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    (b) Moles of CO\u2082 formed<\/strong><\/h4>\n\n\n\n
      \n
    • Mole ratio from the balanced equation<\/strong>:
      2 mol HCl produces 1 mol CO\u2082.<\/li>\n\n\n\n
    • Calculate moles of CO\u2082 produced based on the limiting reactant (HCl)<\/strong>:
      [
      \\text{Moles of CO}_2 = 0.400 \\, \\text{mol HCl} \\times \\frac{1 \\, \\text{mol CO}_2}{2 \\, \\text{mol HCl}} = 0.200 \\, \\text{mol CO}_2
      ]<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      (c) Percent yield<\/strong><\/h4>\n\n\n\n
        \n
      • Actual moles of CO\u2082 formed<\/strong>: 0.180 mol<\/li>\n\n\n\n
      • Theoretical moles of CO\u2082 formed<\/strong>: 0.200 mol<\/li>\n\n\n\n
      • Percent yield calculation<\/strong>:
        [
        \\text{Percent yield} = \\left( \\frac{\\text{Actual yield}}{\\text{Theoretical yield}} \\right) \\times 100 = \\left( \\frac{0.180}{0.200} \\right) \\times 100 = 90\\%
        ]<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

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

        a. Limiting reactant: HCl<\/strong>
        b. Moles of CO\u2082 formed: 0.200 mol<\/strong>
        c. Percent yield: 90%<\/strong><\/p>\n\n\n\n

        \n\n\n\n

        Explanation (300 words):<\/h3>\n\n\n\n

        In this reaction, hydrochloric acid (HCl) reacts with calcium carbonate (CaCO\u2083) to produce carbon dioxide (CO\u2082), water (H\u2082O), and calcium chloride (CaCl\u2082). To solve the problem, the first step is to identify the limiting reactant. The balanced chemical equation shows a 2:1 molar ratio between HCl and CaCO\u2083. With 0.400 mol of HCl and 0.300 mol of CaCO\u2083 available, we calculate that 0.300 mol of CaCO\u2083 would require 0.600 mol of HCl. However, since only 0.400 mol of HCl is present, it limits the reaction.<\/p>\n\n\n\n

        Using the limiting reactant (HCl), we determine how much CO\u2082 is produced. According to the stoichiometry, 2 mol of HCl produces 1 mol of CO\u2082. Therefore, 0.400 mol of HCl generates 0.200 mol of CO\u2082 under ideal conditions (theoretical yield).<\/p>\n\n\n\n

        Finally, the percent yield is calculated to evaluate the reaction efficiency. The actual amount of CO\u2082 formed is 0.180 mol, and dividing this by the theoretical yield (0.200 mol), then multiplying by 100, gives a percent yield of 90%. This indicates that 90% of the expected product was obtained, reflecting some loss or inefficiency in the reaction.<\/p>\n","protected":false},"excerpt":{"rendered":"

        For the reaction of hydrochloric acid and calcium carbonate, the balanced equation is 2 HCl + CaCO3 \u00c3\u00a2\u00e2\u20ac \u00e2\u20ac\u2122 CO2 + H2O + CaCl2 If 0.400 mol HCl is reacted with 0.300 mol CaCO3: a. Which reactant is limiting?\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9___ b. How many moles of CO2 can be formed?\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u00b9___ c. If 0.180 mol of CO2 are […]<\/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-183871","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183871","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=183871"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183871\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183871"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183871"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183871"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}