{"id":185050,"date":"2025-01-22T05:00:28","date_gmt":"2025-01-22T05:00:28","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=185050"},"modified":"2025-01-22T05:00:30","modified_gmt":"2025-01-22T05:00:30","slug":"at-high-temperatures-kcio3-will-decompose-into-kci-and-o2-gas-following-the-reaction-below","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/22\/at-high-temperatures-kcio3-will-decompose-into-kci-and-o2-gas-following-the-reaction-below\/","title":{"rendered":"At high temperatures KCIO3 will decompose into KCI and O2 gas following the reaction below"},"content":{"rendered":"\n

At high temperatures KCIO3 will decompose into KCI and O2 gas following the reaction below. If 1.09 g of KCIO are used, what volume of oxygen gas will be collected at 1.01 atm of pressure and a temperature of 495 \u00b0C? 2KClO3(8)\u2192 2KCl(s) +302 (9)<\/p>\n\n\n\n

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

To solve this, we will use the balanced chemical equation, the ideal gas law, and stoichiometry. Here’s the process:<\/p>\n\n\n\n

Step 1: Balanced Equation<\/h3>\n\n\n\n

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

Step 2: Moles of KClO\u2083<\/h3>\n\n\n\n

The molar mass of (\\text{KClO}_3) is:
[
\\text{KClO}_3 = 39.10 (\\text{K}) + 35.45 (\\text{Cl}) + 3 \\times 16.00 (\\text{O}) = 122.55 \\, \\text{g\/mol}
]<\/p>\n\n\n\n

Given mass of (\\text{KClO}_3 = 1.09 \\, \\text{g}):
[
\\text{Moles of KClO}_3 = \\frac{\\text{mass}}{\\text{molar mass}} = \\frac{1.09}{122.55} = 0.00889 \\, \\text{mol}
]<\/p>\n\n\n\n

Step 3: Moles of O\u2082 Produced<\/h3>\n\n\n\n

From the equation, 2 moles of (\\text{KClO}_3) produce 3 moles of (\\text{O}_2). Thus:
[
\\text{Moles of O}_2 = 0.00889 \\times \\frac{3}{2} = 0.01334 \\, \\text{mol}
]<\/p>\n\n\n\n

Step 4: Ideal Gas Law to Find Volume<\/h3>\n\n\n\n

The ideal gas law is:
[
PV = nRT
]
Rearranged for volume:
[
V = \\frac{nRT}{P}
]<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

    \n
  • (P = 1.01 \\, \\text{atm})<\/li>\n\n\n\n
  • (n = 0.01334 \\, \\text{mol})<\/li>\n\n\n\n
  • (R = 0.0821 \\, \\text{L\u00b7atm\/(mol\u00b7K)})<\/li>\n\n\n\n
  • (T = 495^\\circ\\text{C} = 495 + 273.15 = 768.15 \\, \\text{K})<\/li>\n<\/ul>\n\n\n\n

    [
    V = \\frac{(0.01334)(0.0821)(768.15)}{1.01} = 0.837 \\, \\text{L}
    ]<\/p>\n\n\n\n

    Final Answer:<\/h3>\n\n\n\n

    The volume of oxygen gas collected is approximately 0.837 L<\/strong>.<\/p>\n\n\n\n

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

    This calculation uses stoichiometry to determine the moles of oxygen gas produced from the decomposition of potassium chlorate. The relationship between pressure, volume, temperature, and moles of gas is described by the ideal gas law. The molar ratio from the balanced equation allows conversion from moles of (\\text{KClO}_3) to moles of (\\text{O}_2). The temperature is converted to Kelvin, as required for the ideal gas law. Finally, substituting the known values into the equation yields the volume of gas.<\/p>\n","protected":false},"excerpt":{"rendered":"

    At high temperatures KCIO3 will decompose into KCI and O2 gas following the reaction below. If 1.09 g of KCIO are used, what volume of oxygen gas will be collected at 1.01 atm of pressure and a temperature of 495 \u00b0C? 2KClO3(8)\u2192 2KCl(s) +302 (9) The Correct Answer and Explanation is : To solve this, […]<\/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-185050","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185050","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=185050"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185050\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185050"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185050"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185050"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}