{"id":192547,"date":"2025-02-18T04:33:57","date_gmt":"2025-02-18T04:33:57","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=192547"},"modified":"2025-02-18T04:33:59","modified_gmt":"2025-02-18T04:33:59","slug":"consider-the-decomposition-reaction-of-kcio3-and-answer-the-following-questions","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/18\/consider-the-decomposition-reaction-of-kcio3-and-answer-the-following-questions\/","title":{"rendered":"Consider the decomposition reaction of KCIO3 and answer the following questions"},"content":{"rendered":"\n

Consider the decomposition reaction of KCIO3 and answer the following questions: a. Write the balanced chemical equation of the decomposition reaction of KCIO3. B.Compute for the molar mass of KClO3 and KCl. C. What mass of KCIO3 will produce 48.0 g of oxygen gas, O2, if the decomposition of KCIO3 is complete? D. What mass of KCl is theoretically with 48.0 g of oxygen gas?<\/p>\n\n\n\n

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

A. Balanced Chemical Equation for the Decomposition of Potassium Chlorate (KClO\u2083):<\/h3>\n\n\n\n

The decomposition of potassium chlorate (KClO\u2083) produces potassium chloride (KCl) and oxygen gas (O\u2082). The balanced chemical equation is:<\/p>\n\n\n\n

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

B. Molar Mass of KClO\u2083 and KCl:<\/h3>\n\n\n\n

To compute the molar masses of KClO\u2083 and KCl, we’ll sum up the atomic masses of each element in the compounds.<\/p>\n\n\n\n

    \n
  • Molar Mass of KClO\u2083:<\/strong><\/li>\n\n\n\n
  • Potassium (K): 39.1 g\/mol<\/li>\n\n\n\n
  • Chlorine (Cl): 35.5 g\/mol<\/li>\n\n\n\n
  • Oxygen (O): 16.0 g\/mol \u00d7 3 = 48.0 g\/mol [ \\text{Molar Mass of KClO}_3 = 39.1 + 35.5 + 48.0 = 122.6 \\, \\text{g\/mol} ]<\/li>\n\n\n\n
  • Molar Mass of KCl:<\/strong><\/li>\n\n\n\n
  • Potassium (K): 39.1 g\/mol<\/li>\n\n\n\n
  • Chlorine (Cl): 35.5 g\/mol [ \\text{Molar Mass of KCl} = 39.1 + 35.5 = 74.6 \\, \\text{g\/mol} ]<\/li>\n<\/ul>\n\n\n\n

    C. Mass of KClO\u2083 to Produce 48.0 g of O\u2082:<\/h3>\n\n\n\n

    We are given that 48.0 g of oxygen gas is produced. We can use stoichiometry to find the mass of KClO\u2083 required to produce this amount of oxygen.<\/p>\n\n\n\n

      \n
    1. First, calculate the moles of O\u2082 produced:
      [ \\text{Moles of O}_2 = \\frac{48.0 \\, \\text{g}}{32.0 \\, \\text{g\/mol}} = 1.5 \\, \\text{mol} ]<\/li>\n\n\n\n
    2. From the balanced equation, 2 moles of KClO\u2083 produce 3 moles of O\u2082. So, the moles of KClO\u2083 needed to produce 1.5 moles of O\u2082 is:
      [ \\frac{2 \\, \\text{mol KClO}_3}{3 \\, \\text{mol O}_2} \\times 1.5 \\, \\text{mol O}_2 = 1.0 \\, \\text{mol KClO}_3 ]<\/li>\n\n\n\n
    3. Now, calculate the mass of KClO\u2083 required:
      [ \\text{Mass of KClO}_3 = 1.0 \\, \\text{mol} \\times 122.6 \\, \\text{g\/mol} = 122.6 \\, \\text{g} ]<\/li>\n<\/ol>\n\n\n\n

      D. Mass of KCl Theoretically Produced:<\/h3>\n\n\n\n

      From the balanced equation, for every 2 moles of KClO\u2083 decomposed, 2 moles of KCl are produced. Since we calculated that 1.0 mole of KClO\u2083 is required to produce 1.5 moles of O\u2082, it will also produce 1.0 mole of KCl.<\/p>\n\n\n\n

      Thus, the mass of KCl produced is:
      [ \\text{Mass of KCl} = 1.0 \\, \\text{mol} \\times 74.6 \\, \\text{g\/mol} = 74.6 \\, \\text{g} ]<\/p>\n\n\n\n

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

      In the decomposition of KClO\u2083, the reaction produces potassium chloride (KCl) and oxygen gas (O\u2082). The stoichiometric relationship between the reactants and products allows us to calculate the masses involved in the reaction. By using the molar masses of KClO\u2083 and O\u2082, and the balanced equation, we can find how much KClO\u2083 is required to produce a specific amount of oxygen and how much KCl will theoretically be produced.<\/p>\n\n\n\n

      Let me generate an image of the balanced equation for you.<\/p>\n\n\n\n

      Here is a visual representation of the balanced chemical equation for the decomposition of potassium chlorate (KClO\u2083). Let me know if you’d like any further explanation or have more questions!<\/p>\n\n\n\n

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

      Consider the decomposition reaction of KCIO3 and answer the following questions: a. Write the balanced chemical equation of the decomposition reaction of KCIO3. B.Compute for the molar mass of KClO3 and KCl. C. What mass of KCIO3 will produce 48.0 g of oxygen gas, O2, if the decomposition of KCIO3 is complete? D. What mass […]<\/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-192547","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192547","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=192547"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192547\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=192547"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=192547"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=192547"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}