To solve this problem, we will go step by step for both parts.<\/p>\n\n\n\n
a. Balanced Chemical Equation:<\/h3>\n\n\n\n
The decomposition reaction of potassium chlorate (KClO\u2083) to produce potassium chloride (KCl) and oxygen gas (O\u2082) is represented by the following balanced chemical equation:2KClO3(s)\u21922KCl(s)+3O2(g)2KClO\u2083(s) \\rightarrow 2KCl(s) + 3O\u2082(g)2KClO3\u200b(s)\u21922KCl(s)+3O2\u200b(g)<\/p>\n\n\n\n
This shows that 2 moles of KClO\u2083 decompose to form 2 moles of KCl and 3 moles of oxygen gas.<\/p>\n\n\n\n
b. Mass of KClO\u2083 required to produce 126 L of oxygen gas:<\/h3>\n\n\n\n
We can use the Ideal Gas Law equation to calculate the mass of KClO\u2083 required.<\/p>\n\n\n\n
The Ideal Gas Law is:PV=nRTPV = nRTPV=nRT<\/p>\n\n\n\n
Where:<\/p>\n\n\n\n
- \n
- P<\/strong> = pressure in atm<\/li>\n\n\n\n
- V<\/strong> = volume in liters (L)<\/li>\n\n\n\n
- n<\/strong> = number of moles of gas<\/li>\n\n\n\n
- R<\/strong> = ideal gas constant (0.0821 L\u00b7atm \/ mol\u00b7K)<\/li>\n\n\n\n
- T<\/strong> = temperature in Kelvin (K)<\/li>\n<\/ul>\n\n\n\n
Step 1: Convert the given temperature to Kelvin.<\/h4>\n\n\n\n
Given temperature = 133\u00b0CT(K)=133+273=406 KT(K) = 133 + 273 = 406 \\text{ K}T(K)=133+273=406 K<\/p>\n\n\n\n
Step 2: Rearrange the Ideal Gas Law to find the number of moles of O\u2082.<\/h4>\n\n\n\n
n=PVRTn = \\frac{PV}{RT}n=RTPV\u200b<\/p>\n\n\n\n
Substitute the given values:<\/p>\n\n\n\n
- \n
- P<\/strong> = 0.880 atm<\/li>\n\n\n\n
- V<\/strong> = 126 L<\/li>\n\n\n\n
- R<\/strong> = 0.0821 L\u00b7atm \/ mol\u00b7K<\/li>\n\n\n\n
- T<\/strong> = 406 K<\/li>\n<\/ul>\n\n\n\n
n=(0.880)(126)(0.0821)(406)=110.8833.357\u22483.32 mol O\u2082n = \\frac{(0.880)(126)}{(0.0821)(406)} = \\frac{110.88}{33.357} \\approx 3.32 \\text{ mol O\u2082}n=(0.0821)(406)(0.880)(126)\u200b=33.357110.88\u200b\u22483.32 mol O\u2082<\/p>\n\n\n\n
Step 3: Use stoichiometry to relate moles of O\u2082 to moles of KClO\u2083.<\/h4>\n\n\n\n
From the balanced chemical equation, we know that 3 moles of O\u2082 are produced by 2 moles of KClO\u2083. Therefore:moles of KClO\u2083=23\u00d73.32=2.21 mol KClO\u2083\\text{moles of KClO\u2083} = \\frac{2}{3} \\times 3.32 = 2.21 \\text{ mol KClO\u2083}moles of KClO\u2083=32\u200b\u00d73.32=2.21 mol KClO\u2083<\/p>\n\n\n\n
Step 4: Convert moles of KClO\u2083 to grams.<\/h4>\n\n\n\n
To convert moles of KClO\u2083 to grams, use the molar mass of KClO\u2083. The molar mass of KClO\u2083 is:M(KClO3)=39.1+35.5+(3\u00d716)=122.6 g\/molM(KClO\u2083) = 39.1 + 35.5 + (3 \\times 16) = 122.6 \\text{ g\/mol}M(KClO3\u200b)=39.1+35.5+(3\u00d716)=122.6 g\/mol<\/p>\n\n\n\n
Now, calculate the mass of KClO\u2083:mass of KClO\u2083=2.21 mol\u00d7122.6 g\/mol\u2248270.5 g\\text{mass of KClO\u2083} = 2.21 \\text{ mol} \\times 122.6 \\text{ g\/mol} \\approx 270.5 \\text{ g}mass of KClO\u2083=2.21 mol\u00d7122.6 g\/mol\u2248270.5 g<\/p>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
The mass of KClO\u2083 required to produce 126 L of oxygen gas at 133\u00b0C and 0.880 atm is approximately 270.5 g<\/strong>.<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-10.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Decomposition reaction of KClO3 can be used to produce an oxygen gas. (The other reaction product is solid KCl). a. Write a balanced chemical equation for this reaction. b. What mass of KClO3 must be decomposed to produce 126 L of oxygen gas at 133oC and 0.880 atm? The Correct Answer and Explanation is: To […]<\/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-239221","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239221","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=239221"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239221\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=239221"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=239221"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=239221"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- V<\/strong> = 126 L<\/li>\n\n\n\n
- V<\/strong> = volume in liters (L)<\/li>\n\n\n\n