{"id":190330,"date":"2025-02-12T06:27:06","date_gmt":"2025-02-12T06:27:06","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=190330"},"modified":"2025-02-12T06:27:08","modified_gmt":"2025-02-12T06:27:08","slug":"in-the-decomposition-reaction","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/12\/in-the-decomposition-reaction\/","title":{"rendered":"In the decomposition reaction"},"content":{"rendered":"\n
    \n
  1. In the decomposition reaction: 2KClO3(s)——-> 2KCl(s) + 3O2(g), 3.66 g of KClO3 is decomposing.<\/li>\n<\/ol>\n\n\n\n

    a. What volume of O2 is collected at STP?<\/p>\n\n\n\n

    b. What volume of O2 is collected at 30.0 degree C and 745mmHg?<\/p>\n\n\n\n

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

    Let’s break down the problem step by step, and then I’ll provide the necessary explanation.<\/p>\n\n\n\n

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

    The given reaction is:
    [ 2 \\text{KClO}_3 (s) \\rightarrow 2 \\text{KCl} (s) + 3 \\text{O}_2 (g) ]<\/p>\n\n\n\n

    Given Data:<\/h3>\n\n\n\n
      \n
    • Mass of ( \\text{KClO}_3 ) = 3.66 g<\/li>\n\n\n\n
    • Molar mass of ( \\text{KClO}_3 ) = 122.55 g\/mol<\/li>\n\n\n\n
    • Standard Temperature and Pressure (STP): ( 0^\\circ C ) and ( 1 \\, \\text{atm} ) (which equals ( 22.4 \\, \\text{L\/mol} ) for gases)<\/li>\n\n\n\n
    • Given conditions for second part: 30.0\u00b0C and 745 mmHg<\/li>\n<\/ul>\n\n\n\n

      Part a: Volume of ( O_2 ) Collected at STP<\/h4>\n\n\n\n
        \n
      1. Find moles of ( \\text{KClO}_3 ):<\/strong>
        [
        \\text{Moles of KClO}_3 = \\frac{\\text{Mass}}{\\text{Molar Mass}} = \\frac{3.66 \\, \\text{g}}{122.55 \\, \\text{g\/mol}} = 0.0298 \\, \\text{mol}
        ]<\/li>\n\n\n\n
      2. Use stoichiometry to find moles of ( O_2 ):<\/strong>
        From the balanced equation, 2 moles of ( \\text{KClO}_3 ) produce 3 moles of ( O_2 ).
        [
        \\text{Moles of O}_2 = \\left( \\frac{3}{2} \\right) \\times 0.0298 \\, \\text{mol} = 0.0447 \\, \\text{mol}
        ]<\/li>\n\n\n\n
      3. Use the molar volume at STP to find the volume of ( O_2 ):<\/strong>
        At STP, 1 mole of any ideal gas occupies 22.4 L.
        [
        \\text{Volume of O}_2 = 0.0447 \\, \\text{mol} \\times 22.4 \\, \\text{L\/mol} = 1.00 \\, \\text{L}
        ]<\/li>\n<\/ol>\n\n\n\n

        Thus, the volume of ( O_2 ) collected at STP is 1.00 L<\/strong>.<\/p>\n\n\n\n

        Part b: Volume of ( O_2 ) Collected at 30.0\u00b0C and 745 mmHg<\/h4>\n\n\n\n
          \n
        1. Use the ideal gas law to calculate the volume at the given conditions:<\/strong><\/li>\n<\/ol>\n\n\n\n

          The ideal gas law equation is:
          [
          PV = nRT
          ]
          Where:<\/p>\n\n\n\n

            \n
          • ( P ) = pressure in atm = ( \\frac{745 \\, \\text{mmHg}}{760 \\, \\text{mmHg\/atm}} = 0.980 \\, \\text{atm} )<\/li>\n\n\n\n
          • ( V ) = volume of ( O_2 ) (to be determined)<\/li>\n\n\n\n
          • ( n ) = moles of ( O_2 ) = 0.0447 mol<\/li>\n\n\n\n
          • ( R ) = ideal gas constant = 0.0821 L\u00b7atm\/mol\u00b7K<\/li>\n\n\n\n
          • ( T ) = temperature in Kelvin = ( 30.0 + 273.15 = 303.15 \\, \\text{K} )<\/li>\n<\/ul>\n\n\n\n
              \n
            1. Rearrange the ideal gas law to solve for ( V ):<\/strong>
              [
              V = \\frac{nRT}{P}
              ]
              Substituting in the values:
              [
              V = \\frac{(0.0447 \\, \\text{mol}) (0.0821 \\, \\text{L\u00b7atm\/mol\u00b7K}) (303.15 \\, \\text{K})}{0.980 \\, \\text{atm}} = 1.16 \\, \\text{L}
              ]<\/li>\n<\/ol>\n\n\n\n

              Thus, the volume of ( O_2 ) collected at 30.0\u00b0C and 745 mmHg is 1.16 L<\/strong>.<\/p>\n\n\n\n

              Summary:<\/h3>\n\n\n\n
                \n
              • a)<\/strong> The volume of ( O_2 ) collected at STP is 1.00 L<\/strong>.<\/li>\n\n\n\n
              • b)<\/strong> The volume of ( O_2 ) collected at 30.0\u00b0C and 745 mmHg is 1.16 L<\/strong>.<\/li>\n<\/ul>\n\n\n\n

                Now, I\u2019ll generate a visual representation of the key data and the calculations.<\/p>\n\n\n\n

                Here is the visual representation of the volume of ( O_2 ) collected under the two different conditions: STP and 30\u00b0C at 745 mmHg. As you can see, the volume increases slightly when the temperature rises and the pressure is reduced.<\/p>\n\n\n\n

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

                a. What volume of O2 is collected at STP? b. What volume of O2 is collected at 30.0 degree C and 745mmHg? The Correct Answer and Explanation is : Let’s break down the problem step by step, and then I’ll provide the necessary explanation. Decomposition Reaction: The given reaction is:[ 2 \\text{KClO}_3 (s) \\rightarrow 2 […]<\/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-190330","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190330","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=190330"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190330\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190330"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190330"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190330"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}