{"id":239217,"date":"2025-07-02T19:04:57","date_gmt":"2025-07-02T19:04:57","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=239217"},"modified":"2025-07-02T19:04:59","modified_gmt":"2025-07-02T19:04:59","slug":"oxygen-gas-o2-can-be-obtained-from-the-decomposition-reaction-of-potassium-chlorate","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/02\/oxygen-gas-o2-can-be-obtained-from-the-decomposition-reaction-of-potassium-chlorate\/","title":{"rendered":"Oxygen gas (O2) can be obtained from the decomposition reaction of potassium chlorate"},"content":{"rendered":"\n

Oxygen gas (O2) can be obtained from the decomposition reaction of potassium chlorate (KClO3): 2KClO3(s) \u00e2\u2020\u2019 2KCl(s) + 3O2(g) a) How many moles of O2 can be produced by the decomposition of 32.8 g KClO3? b) How many grams of KClO3 are required to decompose to produce 50.0 g O2<\/p>\n\n\n\n

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

) How many moles of O2 can be produced by the decomposition of 32.8 g KClO3?<\/h3>\n\n\n\n

First, we need to use the molar mass of potassium chlorate (KClO3) and the stoichiometry of the reaction to find the number of moles of oxygen gas (O2) produced.<\/p>\n\n\n\n

The balanced decomposition reaction is:<\/p>\n\n\n\n

2\u2009KClO3(s)\u21922\u2009KCl(s)+3\u2009O2(g)2 \\, \\text{KClO}_3(s) \\rightarrow 2 \\, \\text{KCl}(s) + 3 \\, \\text{O}_2(g)2KClO3\u200b(s)\u21922KCl(s)+3O2\u200b(g)<\/p>\n\n\n\n

Step 1: Molar Mass of KClO3<\/h4>\n\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<\/li>\n<\/ul>\n\n\n\n

    Molar mass of KClO3 = 39.1 + 35.5 + (3 \u00d7 16.0) = 122.6 g\/mol<\/p>\n\n\n\n

    Step 2: Convert 32.8 g KClO3 to moles<\/h4>\n\n\n\n

    Moles of KClO3=mass of KClO3molar mass of KClO3=32.8\u2009g122.6\u2009g\/mol=0.267\u2009mol KClO3\\text{Moles of KClO}_3 = \\frac{\\text{mass of KClO}_3}{\\text{molar mass of KClO}_3} = \\frac{32.8 \\, \\text{g}}{122.6 \\, \\text{g\/mol}} = 0.267 \\, \\text{mol KClO}_3Moles of KClO3\u200b=molar mass of KClO3\u200bmass of KClO3\u200b\u200b=122.6g\/mol32.8g\u200b=0.267mol KClO3\u200b<\/p>\n\n\n\n

    Step 3: Use Stoichiometry to Find Moles of O2<\/h4>\n\n\n\n

    From the balanced equation, 2 moles of KClO3 produce 3 moles of O2. So, for every 2 moles of KClO3, we get 3 moles of O2. Therefore, we can use the ratio to find the moles of O2:Moles of O2=0.267\u2009mol KClO3\u00d73\u2009mol O22\u2009mol KClO3=0.4005\u2009mol O2\\text{Moles of O}_2 = 0.267 \\, \\text{mol KClO}_3 \\times \\frac{3 \\, \\text{mol O}_2}{2 \\, \\text{mol KClO}_3} = 0.4005 \\, \\text{mol O}_2Moles of O2\u200b=0.267mol KClO3\u200b\u00d72mol KClO3\u200b3mol O2\u200b\u200b=0.4005mol O2\u200b<\/p>\n\n\n\n

    Thus, 0.4005 moles of O2<\/strong> can be produced from 32.8 g of KClO3.<\/p>\n\n\n\n

    \n\n\n\n

    b) How many grams of KClO3 are required to decompose to produce 50.0 g O2?<\/h3>\n\n\n\n

    Step 1: Convert Grams of O2 to Moles<\/h4>\n\n\n\n

    First, we calculate the number of moles of oxygen gas (O2) that correspond to 50.0 g:<\/p>\n\n\n\n

    Molar mass of O2 = 32.0 g\/mol.Moles of O2=50.0\u2009g O232.0\u2009g\/mol=1.5625\u2009mol O2\\text{Moles of O}_2 = \\frac{50.0 \\, \\text{g O}_2}{32.0 \\, \\text{g\/mol}} = 1.5625 \\, \\text{mol O}_2Moles of O2\u200b=32.0g\/mol50.0g O2\u200b\u200b=1.5625mol O2\u200b<\/p>\n\n\n\n

    Step 2: Use Stoichiometry to Find Moles of KClO3<\/h4>\n\n\n\n

    From the balanced reaction, 3 moles of O2 are produced from 2 moles of KClO3. Therefore, we can use the ratio to find the moles of KClO3 required:Moles of KClO3=1.5625\u2009mol O2\u00d72\u2009mol KClO33\u2009mol O2=1.0417\u2009mol KClO3\\text{Moles of KClO}_3 = 1.5625 \\, \\text{mol O}_2 \\times \\frac{2 \\, \\text{mol KClO}_3}{3 \\, \\text{mol O}_2} = 1.0417 \\, \\text{mol KClO}_3Moles of KClO3\u200b=1.5625mol O2\u200b\u00d73mol O2\u200b2mol KClO3\u200b\u200b=1.0417mol KClO3\u200b<\/p>\n\n\n\n

    Step 3: Convert Moles of KClO3 to Grams<\/h4>\n\n\n\n

    Finally, we convert the moles of KClO3 to grams using the molar mass of KClO3 (122.6 g\/mol):Mass of KClO3=1.0417\u2009mol KClO3\u00d7122.6\u2009g\/mol=127.5\u2009g KClO3\\text{Mass of KClO}_3 = 1.0417 \\, \\text{mol KClO}_3 \\times 122.6 \\, \\text{g\/mol} = 127.5 \\, \\text{g KClO}_3Mass of KClO3\u200b=1.0417mol KClO3\u200b\u00d7122.6g\/mol=127.5g KClO3\u200b<\/p>\n\n\n\n

    Thus, 127.5 grams of KClO3<\/strong> are required to produce 50.0 grams of oxygen gas (O2).<\/p>\n\n\n\n

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

    Oxygen gas (O2) can be obtained from the decomposition reaction of potassium chlorate (KClO3): 2KClO3(s) \u00e2\u2020\u2019 2KCl(s) + 3O2(g) a) How many moles of O2 can be produced by the decomposition of 32.8 g KClO3? b) How many grams of KClO3 are required to decompose to produce 50.0 g O2 The Correct Answer and Explanation […]<\/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-239217","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239217","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=239217"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239217\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=239217"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=239217"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=239217"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}