{"id":218965,"date":"2025-05-24T18:01:40","date_gmt":"2025-05-24T18:01:40","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218965"},"modified":"2025-05-24T18:01:42","modified_gmt":"2025-05-24T18:01:42","slug":"when-heated-kclo3-decomposes-into-kci-and-o2-but-if-this-reaction-produced-36-2g-of-kcl-how-much-o2-was-produced-in-grams","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/24\/when-heated-kclo3-decomposes-into-kci-and-o2-but-if-this-reaction-produced-36-2g-of-kcl-how-much-o2-was-produced-in-grams\/","title":{"rendered":"When heated KCLO3 decomposes into KCI and O2, but if this reaction produced 36.2g of KCL, how much O2 was produced (in grams)"},"content":{"rendered":"\n<p>When heated KCLO3 decomposes into KCI and O2, but if this reaction produced 36.2g of KCL, how much O2 was produced (in grams)?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Given reaction:<\/strong> 2KClO3\u21922KCl+3O22 \\text{KClO}_3 \\rightarrow 2 \\text{KCl} + 3 \\text{O}_2<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate moles of KCl produced<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of KCl:<ul><li>K = 39.1 g\/mol<\/li><li>Cl = 35.45 g\/mol<\/li><\/ul>So, MKCl=39.1+35.45=74.55\u2009g\/molM_{\\text{KCl}} = 39.1 + 35.45 = 74.55 \\, \\text{g\/mol}<\/li>\n\n\n\n<li>Given mass of KCl = 36.2 g<\/li>\n<\/ul>\n\n\n\n<p>moles&nbsp;of&nbsp;KCl=36.2&nbsp;g74.55&nbsp;g\/mol=0.4855&nbsp;mol\\text{moles of KCl} = \\frac{36.2 \\text{ g}}{74.55 \\text{ g\/mol}} = 0.4855 \\text{ mol}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Use stoichiometry to find moles of O\u2082 produced<\/h3>\n\n\n\n<p>From the balanced equation: 2&nbsp;mol&nbsp;KClO3\u21922&nbsp;mol&nbsp;KCl+3&nbsp;mol&nbsp;O22 \\text{ mol KClO}_3 \\rightarrow 2 \\text{ mol KCl} + 3 \\text{ mol O}_2<\/p>\n\n\n\n<p>Mole ratio: 3&nbsp;mol&nbsp;O22&nbsp;mol&nbsp;KCl=1.5&nbsp;mol&nbsp;O2&nbsp;per&nbsp;mol&nbsp;KCl\\frac{3 \\text{ mol O}_2}{2 \\text{ mol KCl}} = 1.5 \\text{ mol O}_2 \\text{ per mol KCl}<\/p>\n\n\n\n<p>So, moles&nbsp;of&nbsp;O2=0.4855\u00d71.5=0.72825&nbsp;mol\\text{moles of } O_2 = 0.4855 \\times 1.5 = 0.72825 \\text{ mol}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate mass of O\u2082 produced<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of O2=2\u00d716.00=32.00\u2009g\/molO_2 = 2 \\times 16.00 = 32.00 \\, \\text{g\/mol}<\/li>\n<\/ul>\n\n\n\n<p>mass&nbsp;of&nbsp;O2=0.72825\u00d732.00=23.304&nbsp;g\\text{mass of } O_2 = 0.72825 \\times 32.00 = 23.304 \\text{ g}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Final answer:<\/strong><\/h2>\n\n\n\n<p>23.3&nbsp;g&nbsp;of&nbsp;O2&nbsp;produced\\boxed{23.3 \\text{ g of } O_2 \\text{ produced}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Explanation <\/h2>\n\n\n\n<p>When potassium chlorate (KClO3\\text{KClO}_3) is heated, it undergoes a decomposition reaction forming potassium chloride (KCl) and oxygen gas (O2\\text{O}_2). The balanced chemical equation is: 2KClO3\u21922KCl+3O22 \\text{KClO}_3 \\rightarrow 2 \\text{KCl} + 3 \\text{O}_2<\/p>\n\n\n\n<p>This tells us that 2 moles of potassium chlorate produce 2 moles of potassium chloride and 3 moles of oxygen gas. The molar relationship between KCl and O\u2082 is therefore 2:3.<\/p>\n\n\n\n<p>Given that 36.2 grams of KCl are formed, the first step is to convert this mass to moles by dividing by the molar mass of KCl (approximately 74.55 g\/mol). This gives us about 0.4855 moles of KCl.<\/p>\n\n\n\n<p>Using the stoichiometric ratio from the balanced equation, for every 2 moles of KCl, 3 moles of O\u2082 are produced. Therefore, the moles of O\u2082 produced is: 0.4855\u2009mol&nbsp;KCl\u00d73\u2009mol&nbsp;O22\u2009mol&nbsp;KCl=0.72825\u2009mol&nbsp;O20.4855 \\, \\text{mol KCl} \\times \\frac{3 \\, \\text{mol O}_2}{2 \\, \\text{mol KCl}} = 0.72825 \\, \\text{mol O}_2<\/p>\n\n\n\n<p>Finally, to find the mass of oxygen gas produced, multiply the moles of oxygen by its molar mass (32.00 g\/mol): 0.72825\u00d732.00=23.3&nbsp;g0.72825 \\times 32.00 = 23.3 \\text{ g}<\/p>\n\n\n\n<p>This calculation assumes the reaction goes to completion with no side reactions or losses. Therefore, the oxygen produced from the decomposition of potassium chlorate that yields 36.2 g of potassium chloride is approximately <strong>23.3 grams<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-399.png\" alt=\"\" class=\"wp-image-218966\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>When heated KCLO3 decomposes into KCI and O2, but if this reaction produced 36.2g of KCL, how much O2 was produced (in grams)? The Correct Answer and Explanation is: Given reaction: 2KClO3\u21922KCl+3O22 \\text{KClO}_3 \\rightarrow 2 \\text{KCl} + 3 \\text{O}_2 Step 1: Calculate moles of KCl produced moles&nbsp;of&nbsp;KCl=36.2&nbsp;g74.55&nbsp;g\/mol=0.4855&nbsp;mol\\text{moles of KCl} = \\frac{36.2 \\text{ g}}{74.55 \\text{ g\/mol}} [&hellip;]<\/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-218965","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218965","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=218965"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218965\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218965"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218965"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218965"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}