{"id":208720,"date":"2025-04-28T05:00:35","date_gmt":"2025-04-28T05:00:35","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=208720"},"modified":"2025-04-28T05:00:37","modified_gmt":"2025-04-28T05:00:37","slug":"45-7-g-of-p4o10-is-reacted-with-an-excess-of-calcium-oxide","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/04\/28\/45-7-g-of-p4o10-is-reacted-with-an-excess-of-calcium-oxide\/","title":{"rendered":"45.7 g of P4o10 is reacted with an excess of calcium oxide"},"content":{"rendered":"\n

45.7 g of P4o10 is reacted with an excess of calcium oxide. what mass of calcium phosphate is produced. ?<\/p>\n\n\n\n

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To solve this problem, let’s start by analyzing the chemical reaction and then work through the calculations step by step.<\/p>\n\n\n\n

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

The reaction between phosphorus pentoxide (P\u2084O\u2081\u2080) and calcium oxide (CaO) produces calcium phosphate (Ca\u2083(PO\u2084)\u2082). The balanced equation for the reaction is:<\/p>\n\n\n\n

[
P_4O_{10} (s) + 6CaO (s) \\rightarrow 2Ca_3(PO_4)_2 (s)
]<\/p>\n\n\n\n

Step 2: Calculate the molar mass of P\u2084O\u2081\u2080<\/h3>\n\n\n\n

The molar mass of phosphorus pentoxide (P\u2084O\u2081\u2080) can be calculated by adding the atomic masses of phosphorus and oxygen:<\/p>\n\n\n\n

    \n
  • Atomic mass of phosphorus (P) = 30.97 g\/mol<\/li>\n\n\n\n
  • Atomic mass of oxygen (O) = 16.00 g\/mol<\/li>\n<\/ul>\n\n\n\n

    The molar mass of P\u2084O\u2081\u2080 is:<\/p>\n\n\n\n

    [
    M_{P_4O_{10}} = 4 \\times 30.97 \\, \\text{g\/mol} + 10 \\times 16.00 \\, \\text{g\/mol}
    ]
    [
    M_{P_4O_{10}} = 123.88 \\, \\text{g\/mol} + 160.00 \\, \\text{g\/mol} = 283.88 \\, \\text{g\/mol}
    ]<\/p>\n\n\n\n

    Step 3: Calculate the moles of P\u2084O\u2081\u2080<\/h3>\n\n\n\n

    You are given 45.7 grams of P\u2084O\u2081\u2080. To find the number of moles, use the formula:<\/p>\n\n\n\n

    [
    \\text{moles of P\u2084O\u2081\u2080} = \\frac{\\text{mass of P\u2084O\u2081\u2080}}{\\text{molar mass of P\u2084O\u2081\u2080}}
    ]
    [
    \\text{moles of P\u2084O\u2081\u2080} = \\frac{45.7 \\, \\text{g}}{283.88 \\, \\text{g\/mol}} = 0.1616 \\, \\text{mol}
    ]<\/p>\n\n\n\n

    Step 4: Use stoichiometry to find the moles of Ca\u2083(PO\u2084)\u2082<\/h3>\n\n\n\n

    From the balanced equation, we know that 1 mole of P\u2084O\u2081\u2080 reacts to produce 2 moles of Ca\u2083(PO\u2084)\u2082. Therefore, the moles of Ca\u2083(PO\u2084)\u2082 produced are:<\/p>\n\n\n\n

    [
    \\text{moles of Ca\u2083(PO\u2084)\u2082} = 2 \\times \\text{moles of P\u2084O\u2081\u2080}
    ]
    [
    \\text{moles of Ca\u2083(PO\u2084)\u2082} = 2 \\times 0.1616 \\, \\text{mol} = 0.3232 \\, \\text{mol}
    ]<\/p>\n\n\n\n

    Step 5: Calculate the mass of Ca\u2083(PO\u2084)\u2082<\/h3>\n\n\n\n

    Now, calculate the molar mass of calcium phosphate (Ca\u2083(PO\u2084)\u2082):<\/p>\n\n\n\n

      \n
    • Atomic mass of calcium (Ca) = 40.08 g\/mol<\/li>\n\n\n\n
    • Atomic mass of phosphorus (P) = 30.97 g\/mol<\/li>\n\n\n\n
    • Atomic mass of oxygen (O) = 16.00 g\/mol<\/li>\n<\/ul>\n\n\n\n

      The molar mass of Ca\u2083(PO\u2084)\u2082 is:<\/p>\n\n\n\n

      [
      M_{\\text{Ca}3(\\text{PO}_4)_2} = 3 \\times 40.08 \\, \\text{g\/mol} + 2 \\times (2 \\times 30.97 \\, \\text{g\/mol} + 8 \\times 16.00 \\, \\text{g\/mol}) ] [ M<\/em>{\\text{Ca}3(\\text{PO}_4)_2} = 120.24 \\, \\text{g\/mol} + 2 \\times (61.94 \\, \\text{g\/mol} + 128.00 \\, \\text{g\/mol}) ] [ M<\/em>{\\text{Ca}_3(\\text{PO}_4)_2} = 120.24 \\, \\text{g\/mol} + 2 \\times 189.94 \\, \\text{g\/mol} = 120.24 \\, \\text{g\/mol} + 379.88 \\, \\text{g\/mol} = 500.12 \\, \\text{g\/mol}
      ]<\/p>\n\n\n\n

      Now, use the number of moles of Ca\u2083(PO\u2084)\u2082 to calculate the mass:<\/p>\n\n\n\n

      [
      \\text{mass of Ca\u2083(PO\u2084)\u2082} = \\text{moles of Ca\u2083(PO\u2084)\u2082} \\times M_{\\text{Ca}_3(\\text{PO}_4)_2}
      ]
      [
      \\text{mass of Ca\u2083(PO\u2084)\u2082} = 0.3232 \\, \\text{mol} \\times 500.12 \\, \\text{g\/mol}
      ]
      [
      \\text{mass of Ca\u2083(PO\u2084)\u2082} = 161.22 \\, \\text{g}
      ]<\/p>\n\n\n\n

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

      The mass of calcium phosphate (Ca\u2083(PO\u2084)\u2082) produced is 161.22 grams<\/strong>.<\/p>\n\n\n\n

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

      This calculation involves determining the amount of calcium phosphate produced from a given amount of phosphorus pentoxide (P\u2084O\u2081\u2080) in excess calcium oxide. We first used stoichiometric relationships from the balanced chemical equation, converting the given mass of P\u2084O\u2081\u2080 into moles and using the molar ratio between P\u2084O\u2081\u2080 and Ca\u2083(PO\u2084)\u2082 to determine the moles of calcium phosphate formed. Finally, we converted the moles of Ca\u2083(PO\u2084)\u2082 into mass using its molar mass. This approach ensures accurate calculation of the product’s mass in a chemical reaction.<\/p>\n","protected":false},"excerpt":{"rendered":"

      45.7 g of P4o10 is reacted with an excess of calcium oxide. what mass of calcium phosphate is produced. ? The correct answer and explanation is : To solve this problem, let’s start by analyzing the chemical reaction and then work through the calculations step by step. Step 1: Write the balanced chemical equation 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-208720","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208720","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=208720"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208720\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=208720"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=208720"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=208720"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}