{"id":190550,"date":"2025-02-12T11:47:20","date_gmt":"2025-02-12T11:47:20","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=190550"},"modified":"2025-02-12T11:47:40","modified_gmt":"2025-02-12T11:47:40","slug":"a-mixture-containing-only-bao-and-cao-weighs-2-00-g","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/12\/a-mixture-containing-only-bao-and-cao-weighs-2-00-g\/","title":{"rendered":"A mixture containing only BaO and CaO weighs 2.00 g"},"content":{"rendered":"\n
    \n
  1. A mixture containing only BaO and CaO weighs 2.00 g. The oxides are converted to the corresponding mixed sulfates, which weigh 4.00 g. Calculate the percent Ba and Ca in the original mixture.<\/li>\n\n\n\n
  2. A mixture containing only BaSO4 and CaSO4 contains one-half as much Ba2+ as Ca2+ by weight. What is the percentage of CaSO4 in the mixture?<\/li>\n<\/ol>\n\n\n\n

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

    Problem 1:<\/h3>\n\n\n\n

    Given:<\/strong><\/p>\n\n\n\n

      \n
    • A mixture of BaO and CaO weighs 2.00 g.<\/li>\n\n\n\n
    • The mixture is converted to the corresponding mixed sulfates, BaSO\u2084 and CaSO\u2084, weighing 4.00 g.<\/li>\n<\/ul>\n\n\n\n

      Objective:<\/strong>
      To calculate the percent of Ba and Ca in the original mixture of BaO and CaO.<\/p>\n\n\n\n

      Solution:<\/h3>\n\n\n\n

      Let the masses of BaO and CaO in the mixture be represented as ( x ) and ( y ), respectively. So,
      [
      x + y = 2.00 \\, \\text{g}
      ]
      The BaO reacts with sulfuric acid to form BaSO\u2084, and similarly, CaO reacts with sulfuric acid to form CaSO\u2084. The molar masses of BaO and CaO are 137.33 g\/mol and 56.08 g\/mol, respectively. The molar masses of BaSO\u2084 and CaSO\u2084 are 233.39 g\/mol and 136.14 g\/mol, respectively.<\/p>\n\n\n\n

        \n
      1. The mass of BaSO\u2084 formed is related to the mass of BaO:
        [
        \\text{Mass of BaSO\u2084} = \\left( \\frac{233.39}{137.33} \\right) \\times x
        ]<\/li>\n\n\n\n
      2. The mass of CaSO\u2084 formed is related to the mass of CaO:
        [
        \\text{Mass of CaSO\u2084} = \\left( \\frac{136.14}{56.08} \\right) \\times y
        ]
        The total mass of the sulfates is 4.00 g:
        [
        \\left( \\frac{233.39}{137.33} \\right) \\times x + \\left( \\frac{136.14}{56.08} \\right) \\times y = 4.00
        ]<\/li>\n<\/ol>\n\n\n\n

        Now, solve this system of equations:<\/p>\n\n\n\n

          \n
        1. ( x + y = 2.00 )<\/li>\n\n\n\n
        2. ( \\left( \\frac{233.39}{137.33} \\right) \\times x + \\left( \\frac{136.14}{56.08} \\right) \\times y = 4.00 )<\/li>\n<\/ol>\n\n\n\n

          From these, you can find ( x ) (mass of BaO) and ( y ) (mass of CaO). Once you have those, you can calculate the percentage of Ba and Ca in the original mixture.<\/p>\n\n\n\n

          \n\n\n\n

          Problem 2:<\/h3>\n\n\n\n

          Given:<\/strong><\/p>\n\n\n\n

            \n
          • A mixture contains BaSO\u2084 and CaSO\u2084.<\/li>\n\n\n\n
          • The mixture contains one-half as much Ba\u00b2\u207a as Ca\u00b2\u207a by weight.<\/li>\n<\/ul>\n\n\n\n

            Objective:<\/strong>
            To find the percentage of CaSO\u2084 in the mixture.<\/p>\n\n\n\n

            Solution:<\/h3>\n\n\n\n

            Let the mass of BaSO\u2084 be ( m_{\\text{BaSO\u2084}} ) and the mass of CaSO\u2084 be ( m_{\\text{CaSO\u2084}} ) in the mixture. The total mass is:
            [
            m_{\\text{BaSO\u2084}} + m_{\\text{CaSO\u2084}} = \\text{Total mass of the mixture}
            ]
            The amount of Ba\u00b2\u207a and Ca\u00b2\u207a in the mixture comes from the dissociation of BaSO\u2084 and CaSO\u2084.<\/p>\n\n\n\n

              \n
            1. The molar mass of BaSO\u2084 is 233.39 g\/mol, and Ba\u00b2\u207a makes up ( \\frac{137.33}{233.39} ) of the molar mass.<\/li>\n\n\n\n
            2. The molar mass of CaSO\u2084 is 136.14 g\/mol, and Ca\u00b2\u207a makes up ( \\frac{40.08}{136.14} ) of the molar mass.<\/li>\n<\/ol>\n\n\n\n

              The total mass of Ba\u00b2\u207a and Ca\u00b2\u207a is related to the masses of BaSO\u2084 and CaSO\u2084:
              [
              \\frac{137.33}{233.39} \\times m_{\\text{BaSO\u2084}} = \\frac{1}{2} \\times \\frac{40.08}{136.14} \\times m_{\\text{CaSO\u2084}}
              ]
              From this, you can solve for the percentage of CaSO\u2084 in the mixture.<\/p>\n\n\n\n

              \n\n\n\n

              I\u2019ll proceed with the calculations for both problems and also generate a relevant image.<\/p>\n\n\n\n

              Problem 1:<\/h3>\n\n\n\n

              The mass of BaO in the mixture is approximately 1.175 g<\/strong>, and the mass of CaO is approximately 0.825 g<\/strong>.<\/p>\n\n\n\n

                \n
              • The percentage of Ba in the original mixture is 58.73%<\/strong>.<\/li>\n\n\n\n
              • The percentage of Ca in the original mixture is 41.27%<\/strong>.<\/li>\n<\/ul>\n\n\n\n

                Problem 2:<\/h3>\n\n\n\n
                  \n
                • The percentage of CaSO\u2084 in the mixture is approximately 79.99%<\/strong>.<\/li>\n<\/ul>\n\n\n\n

                  Explanation:<\/h3>\n\n\n\n

                  For Problem 1, we used a system of equations to calculate the masses of BaO and CaO in the mixture. By using their respective molar mass ratios with BaSO\u2084 and CaSO\u2084, we were able to find the masses of each oxide in the mixture. The percentage of each component was then calculated based on the total mass of the mixture.<\/p>\n\n\n\n

                  For Problem 2, we applied a mass balance for the Ba\u00b2\u207a and Ca\u00b2\u207a ions in the sulfates, considering that the mass of Ba\u00b2\u207a is half the mass of Ca\u00b2\u207a. Using the molar masses of BaSO\u2084 and CaSO\u2084, we solved for the relative proportions of the sulfates in the mixture, yielding the percentage of CaSO\u2084.<\/p>\n\n\n\n

                  I will now generate an image to visualize the process for both problems.<\/p>\n\n\n\n

                  Here is the visual representation of the process for both problems, showing the conversion of BaO and CaO to BaSO\u2084 and CaSO\u2084, along with the calculated percentages. Let me know if you need further clarification!<\/p>\n\n\n\n

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

                  The Correct Answer and Explanation is : Problem 1: Given: Objective:To calculate the percent of Ba and Ca in the original mixture of BaO and CaO. Solution: Let the masses of BaO and CaO in the mixture be represented as ( x ) and ( y ), respectively. So,[x + y = 2.00 \\, \\text{g}]The […]<\/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-190550","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190550","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=190550"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190550\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190550"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190550"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190550"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}