{"id":206587,"date":"2025-04-02T15:13:22","date_gmt":"2025-04-02T15:13:22","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=206587"},"modified":"2025-04-02T15:13:25","modified_gmt":"2025-04-02T15:13:25","slug":"decomposition-of-potassium-chlorate-kclo3-produces-potassium-chloride-kcl-and-pure-oxygen-o2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/04\/02\/decomposition-of-potassium-chlorate-kclo3-produces-potassium-chloride-kcl-and-pure-oxygen-o2\/","title":{"rendered":"Decomposition of potassium chlorate (KClO3) produces potassium chloride (KCl) and pure oxygen (O2)"},"content":{"rendered":"\n<p>Decomposition of potassium chlorate (KClO3) produces potassium chloride (KCl) and pure oxygen (O2). The balanced equation for the reaction is as follows: 2KClO3(s) &#8212;&gt; 2KCl(s) + 3O2(g). What volume of oxygen gas is released at STP if 10.0 g of potassium chlorate is decomposed? (The molar mass of KClO3 is 122.55 g\/mol.)<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To determine the volume of oxygen gas (O\u2082) released at standard temperature and pressure (STP) when 10.0 g of potassium chlorate (KClO\u2083) decomposes, follow these steps:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the Moles of KClO\u2083<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of KClO\u2083 = 10.0 g<\/li>\n\n\n\n<li>Molar mass of KClO\u2083 = 122.55 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Moles of KClO\u2083:<br>[<br>\\frac{10.0 \\text{ g}}{122.55 \\text{ g\/mol}} = 0.0816 \\text{ moles}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Use the Balanced Equation to Find Moles of O\u2082<\/h3>\n\n\n\n<p>From the reaction:<br>[<br>2KClO\u2083 \\rightarrow 2KCl + 3O\u2082<br>]<br>The molar ratio of KClO\u2083 to O\u2082 is <strong>2:3<\/strong>. Using this ratio:<\/p>\n\n\n\n<p>[<br>0.0816 \\text{ moles KClO\u2083} \\times \\frac{3 \\text{ moles O\u2082}}{2 \\text{ moles KClO\u2083}}<br>]<\/p>\n\n\n\n<p>[<br>= 0.1224 \\text{ moles O\u2082}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Convert Moles of O\u2082 to Volume at STP<\/h3>\n\n\n\n<p>At STP, <strong>1 mole of any gas occupies 22.4 L<\/strong>.<br>[<br>0.1224 \\text{ moles O\u2082} \\times 22.4 \\text{ L\/mole}<br>]<\/p>\n\n\n\n<p>[<br>= 2.74 \\text{ L O\u2082}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p><strong>The volume of oxygen gas released is 2.74 L at STP.<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>This problem involves stoichiometric calculations using the balanced chemical equation for potassium chlorate decomposition. The key steps include:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Converting mass of reactant (KClO\u2083) to moles using molar mass.<\/li>\n\n\n\n<li>Using the balanced equation to find the stoichiometric relationship between KClO\u2083 and O\u2082.<\/li>\n\n\n\n<li>Converting moles of O\u2082 to volume using the molar volume of a gas at STP.<\/li>\n<\/ol>\n\n\n\n<p>This method ensures accurate results by following a structured approach to chemical calculations.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img decoding=\"async\" src=\"https:\/\/files.oaiusercontent.com\/file-RFbdT7GtRJg1qF56vfUnZV?se=2025-04-02T15%3A15%3A56Z&amp;sp=r&amp;sv=2024-08-04&amp;sr=b&amp;rscc=max-age%3D604800%2C%20immutable%2C%20private&amp;rscd=attachment%3B%20filename%3Dac3bfa09-2f1b-466e-a24a-5b2f2e756bed.webp&amp;sig=AwaInyT4VH%2BK1kZPczcDiJqILSsSTRSd5EAsHeDlc\/o%3D\" alt=\"\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Decomposition of potassium chlorate (KClO3) produces potassium chloride (KCl) and pure oxygen (O2). The balanced equation for the reaction is as follows: 2KClO3(s) &#8212;&gt; 2KCl(s) + 3O2(g). What volume of oxygen gas is released at STP if 10.0 g of potassium chlorate is decomposed? (The molar mass of KClO3 is 122.55 g\/mol.) The correct answer [&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-206587","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/206587","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=206587"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/206587\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=206587"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=206587"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=206587"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}