{"id":38499,"date":"2025-06-26T10:12:46","date_gmt":"2025-06-26T10:12:46","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=38499"},"modified":"2025-06-26T10:12:48","modified_gmt":"2025-06-26T10:12:48","slug":"given-the-following-reaction-caoh2-h2so4-a%e2%80%a0-caso4-2h2o-if-you-start-with-14-82-g-of-caoh2-and-16-35-g-of-h2so4","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/given-the-following-reaction-caoh2-h2so4-a%e2%80%a0-caso4-2h2o-if-you-start-with-14-82-g-of-caoh2-and-16-35-g-of-h2so4\/","title":{"rendered":"Given the following reaction: Ca(OH)2 + H2SO4 \u00e2\u2020\u2019 CaSO4 + 2H2O If you start with 14.82 g of Ca(OH)2 and 16.35 g of H2SO4,"},"content":{"rendered":"\n

Given the following reaction: Ca(OH)2 + H2SO4 \u00e2\u2020\u2019 CaSO4 + 2H2O If you start with 14.82 g of Ca(OH)2 and 16.35 g of H2SO4, a) Determine the limiting reagent. b) Determine the number of moles of H2O produced. c) Determine the number of grams of CaSO4 produced. d) Determine the number of grams of excess reagent left.<\/p>\n\n\n\n

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

To solve this stoichiometry problem, we will follow these steps:<\/p>\n\n\n\n

Balanced Chemical Equation:<\/strong><\/h3>\n\n\n\n

Ca(OH)2+H2SO4\u2192CaSO4+2H2O\\text{Ca(OH)}_2 + \\text{H}_2\\text{SO}_4 \\rightarrow \\text{CaSO}_4 + 2\\text{H}_2\\text{O}Ca(OH)2\u200b+H2\u200bSO4\u200b\u2192CaSO4\u200b+2H2\u200bO<\/p>\n\n\n\n

This equation tells us that 1 mole of calcium hydroxide reacts with 1 mole of sulfuric acid to produce 1 mole of calcium sulfate and 2 moles of water.<\/p>\n\n\n\n

\n\n\n\n

Step 1: Molar Masses<\/strong><\/h3>\n\n\n\n
    \n
  • Ca(OH)\u2082 = 40.08 + (16.00 + 1.008) \u00d7 2 = 74.096 g\/mol<\/strong><\/li>\n\n\n\n
  • H\u2082SO\u2084 = 2(1.008) + 32.06 + 4(16.00) = 98.076 g\/mol<\/strong><\/li>\n\n\n\n
  • CaSO\u2084 = 40.08 + 32.06 + 4(16.00) = 136.14 g\/mol<\/strong><\/li>\n\n\n\n
  • H\u2082O = 2(1.008) + 16.00 = 18.016 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Step 2: Convert Mass to Moles<\/strong><\/h3>\n\n\n\n
      \n
    • Moles of Ca(OH)\u2082 = 14.82 g \u00f7 74.096 g\/mol = 0.2000 mol<\/strong><\/li>\n\n\n\n
    • Moles of H\u2082SO\u2084 = 16.35 g \u00f7 98.076 g\/mol = 0.1667 mol<\/strong><\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Step 3: Identify the Limiting Reagent<\/strong><\/h3>\n\n\n\n

      From the balanced equation, the molar ratio is 1:1.
      Compare the actual moles available:<\/p>\n\n\n\n

        \n
      • Ca(OH)\u2082: 0.2000 mol<\/li>\n\n\n\n
      • H\u2082SO\u2084: 0.1667 mol<\/li>\n<\/ul>\n\n\n\n

        Since H\u2082SO\u2084 has fewer moles than Ca(OH)\u2082, H\u2082SO\u2084 is the limiting reagent<\/strong>.<\/p>\n\n\n\n

        \n\n\n\n

        Step 4: Calculate Moles of H\u2082O Produced<\/strong><\/h3>\n\n\n\n

        From the equation: 1 mol H\u2082SO\u2084 \u2192 2 mol H\u2082O
        So, 0.1667 mol H\u2082SO\u2084 \u2192 0.1667 \u00d7 2 = 0.3334 mol H\u2082O<\/strong><\/p>\n\n\n\n

        \n\n\n\n

        Step 5: Calculate Grams of CaSO\u2084 Produced<\/strong><\/h3>\n\n\n\n

        From the equation: 1 mol H\u2082SO\u2084 \u2192 1 mol CaSO\u2084
        So, 0.1667 mol CaSO\u2084 \u00d7 136.14 g\/mol = 22.69 g CaSO\u2084<\/strong><\/p>\n\n\n\n

        \n\n\n\n

        Step 6: Calculate Excess Reagent Left (Ca(OH)\u2082)<\/strong><\/h3>\n\n\n\n

        Used Ca(OH)\u2082 = 0.1667 mol
        Initial = 0.2000 mol
        Remaining = 0.2000 – 0.1667 = 0.0333 mol
        Mass = 0.0333 mol \u00d7 74.096 g\/mol = 2.47 g Ca(OH)\u2082 left<\/strong><\/p>\n\n\n\n

        \n\n\n\n

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

        a) Limiting reagent: H\u2082SO\u2084<\/strong>
        b) Moles of H\u2082O produced: 0.3334 mol<\/strong>
        c) Grams of CaSO\u2084 produced: 22.69 g<\/strong>
        d) Grams of excess Ca(OH)\u2082 left: 2.47 g<\/strong><\/p>\n\n\n\n

        \n\n\n\n

        Explanation <\/strong><\/h3>\n\n\n\n

        This chemical reaction between calcium hydroxide and sulfuric acid is a typical acid-base reaction. To determine the limiting reagent, we first need to calculate the moles of each reactant by dividing their given masses by their molar masses. The balanced equation shows a 1:1 mole ratio between the two reactants, so we directly compare the available moles. The reactant with fewer moles is the limiting reagent because it will be completely consumed first, stopping the reaction.<\/p>\n\n\n\n

        In this case, sulfuric acid is the limiting reagent since we only have 0.1667 moles compared to 0.2000 moles of calcium hydroxide. Therefore, all calculations for product formation are based on the amount of sulfuric acid available.<\/p>\n\n\n\n

        Using stoichiometry, we find that every mole of sulfuric acid produces two moles of water. Thus, 0.1667 moles of sulfuric acid yield 0.3334 moles of water. To find the mass of calcium sulfate formed, we multiply the number of moles of CaSO\u2084 (equal to the moles of H\u2082SO\u2084 used) by its molar mass.<\/p>\n\n\n\n

        Lastly, to determine how much calcium hydroxide remains, we subtract the amount that reacted from the initial amount and convert the remaining moles back into grams. This provides the mass of the excess reagent left over after the reaction.<\/p>\n\n\n\n

        By following this structured approach, we accurately determine the limiting reagent, the amount of water and calcium sulfate produced, and how much excess calcium hydroxide remains.<\/p>\n\n\n\n

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

        Given the following reaction: Ca(OH)2 + H2SO4 \u00e2\u2020\u2019 CaSO4 + 2H2O If you start with 14.82 g of Ca(OH)2 and 16.35 g of H2SO4, a) Determine the limiting reagent. b) Determine the number of moles of H2O produced. c) Determine the number of grams of CaSO4 produced. d) Determine the number of grams of excess […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-38499","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38499","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=38499"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38499\/revisions"}],"predecessor-version":[{"id":38501,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38499\/revisions\/38501"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=38499"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=38499"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=38499"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}