{"id":218510,"date":"2025-05-23T21:50:17","date_gmt":"2025-05-23T21:50:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218510"},"modified":"2025-05-23T21:50:19","modified_gmt":"2025-05-23T21:50:19","slug":"what-mass-of-ko2-in-grams-is-required-to-react-with-15-9-l-of-co2-at-23-0c-and-955-mm-hg-_-g-bc","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/23\/what-mass-of-ko2-in-grams-is-required-to-react-with-15-9-l-of-co2-at-23-0c-and-955-mm-hg-_-g-bc\/","title":{"rendered":"What mass of KO2, in grams, is required to react with 15.9 L of CO2 at 23.0\u00b0C and 955 mm Hg? _ g (b)(c)"},"content":{"rendered":"\n

(a)What mass of KO2, in grams, is required to react with 15.9 L of CO2 at 23.0\u00b0C and 955 mm Hg? _<\/em> g (b)(c)<\/p>\n\n\n\n

A self-contained underwater breathing apparatus uses canisters containing potassium superoxide. The superoxide consumes the exhaled CO2 by a person and replaces it with oxygen.<\/p>\n\n\n\n

4KO2 (s) + 2CO2 (g) —> 2K2CO3 (s) + 3 O2 (g)<\/p>\n\n\n\n

Hydrazine reacts with O2 according to the following equation: N2H4 (g) + O2 (g) —> N2 (g) + 2H2O (l)<\/p>\n\n\n\n

Assume the O2 needed for the reaction is in a 650. L tank at 16.0 \u00b0C. What must the oxygen pressure be in the tank to have enough oxygen to consume 1.28 kg of hydrazine completely? ____atm<\/p>\n\n\n\n

A 0.0125-g sample of a gas with an empirical formula of CHF2 is placed in a 165-mL flask. It has a pressure of 6.85 mm Hg at 22.5 \u00b0C. What is the molecular formula of the compound?<\/p>\n\n\n\n

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

Let’s go through each part of the question in sequence.<\/p>\n\n\n\n

\n\n\n\n

(a) What mass of KO\u2082 is required to react with 15.9 L of CO\u2082 at 23.0\u202f\u00b0C and 955 mm Hg?<\/strong><\/h3>\n\n\n\n

Given reaction:<\/strong> 4KO2(s)+2CO2(g)\u21922K2CO3(s)+3O2(g)4KO_2 (s) + 2CO_2 (g) \\rightarrow 2K_2CO_3 (s) + 3O_2 (g)<\/p>\n\n\n\n

We first calculate the number of moles of CO\u2082 using the ideal gas law<\/strong>: PV=nRTPV = nRT<\/p>\n\n\n\n

Convert units first:<\/strong><\/p>\n\n\n\n

    \n
  • P=955\u00a0mmHg=955760\u00a0atm\u22481.2566\u00a0atmP = 955\\ \\text{mmHg} = \\frac{955}{760}\\ \\text{atm} \\approx 1.2566\\ \\text{atm}<\/li>\n\n\n\n
  • V=15.9\u00a0LV = 15.9\\ \\text{L}<\/li>\n\n\n\n
  • T=23.0\u00a0\u2218C=296.15\u00a0KT = 23.0\\ ^\\circ C = 296.15\\ \\text{K}<\/li>\n\n\n\n
  • R=0.0821\u00a0L\\cdotpatm\/mol\\cdotpKR = 0.0821\\ \\text{L\u00b7atm\/mol\u00b7K}<\/li>\n<\/ul>\n\n\n\n

    n=PVRT=(1.2566)(15.9)(0.0821)(296.15)\u22480.8215 mol CO\u2082n = \\frac{PV}{RT} = \\frac{(1.2566)(15.9)}{(0.0821)(296.15)} \\approx 0.8215\\ \\text{mol CO\u2082}<\/p>\n\n\n\n

    From the reaction:
    2 mol CO\u2082 \u2192 4 mol KO\u2082 \u2192 mol ratio = 2:4 = 1:2<\/strong> mol KO\u2082=0.8215\u00d72=1.643 mol\\text{mol KO\u2082} = 0.8215 \\times 2 = 1.643\\ \\text{mol}<\/p>\n\n\n\n

    Molar mass of KO\u2082 = 39.10 (K) + 2\u00d716.00 (O) = 71.10 g\/mol<\/strong> Mass KO\u2082=1.643\u00d771.10\u2248116.8 g\\text{Mass KO\u2082} = 1.643 \\times 71.10 \\approx \\boxed{116.8\\ \\text{g}}<\/p>\n\n\n\n

    \n\n\n\n

    (b) What must the oxygen pressure be in the tank to consume 1.28 kg of hydrazine?<\/strong><\/h3>\n\n\n\n

    Reaction: N\u2082H\u2084 (g) + O\u2082 (g) \u2192 N\u2082 (g) + 2H\u2082O (l)\\text{N\u2082H\u2084 (g) + O\u2082 (g) \u2192 N\u2082 (g) + 2H\u2082O (l)}<\/p>\n\n\n\n

    Step 1:<\/strong> Moles of N\u2082H\u2084:<\/p>\n\n\n\n

    Molar mass N\u2082H\u2084 = 2\u00d714.01 + 4\u00d71.008 = 32.05 g\/mol<\/strong> Mass=1.28 kg=1280 g, mol=128032.05\u224839.93 mol\\text{Mass} = 1.28\\ \\text{kg} = 1280\\ \\text{g},\\ \\text{mol} = \\frac{1280}{32.05} \\approx 39.93\\ \\text{mol}<\/p>\n\n\n\n

    From the equation: 1 mol N\u2082H\u2084 reacts with 1 mol O\u2082 \u2192 mol O\u2082 = 39.93 mol<\/p>\n\n\n\n

    Step 2:<\/strong> Use PV = nRT to find pressure.<\/p>\n\n\n\n

      \n
    • V=650.0\u00a0LV = 650.0\\ \\text{L}<\/li>\n\n\n\n
    • T=16.0\u00b0C=289.15\u00a0KT = 16.0\u00b0C = 289.15\\ \\text{K}<\/li>\n\n\n\n
    • R=0.0821R = 0.0821<\/li>\n<\/ul>\n\n\n\n

      P=nRTV=(39.93)(0.0821)(289.15)650.0\u22481.46 atmP = \\frac{nRT}{V} = \\frac{(39.93)(0.0821)(289.15)}{650.0} \\approx \\boxed{1.46\\ \\text{atm}}<\/p>\n\n\n\n

      \n\n\n\n

      (c) Determine the molecular formula of a gas with empirical formula CHF\u2082<\/strong><\/h3>\n\n\n\n

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

        \n
      • Mass = 0.0125 g<\/li>\n\n\n\n
      • Volume = 165 mL = 0.165 L<\/li>\n\n\n\n
      • Pressure = 6.85 mmHg = 6.85 \/ 760 = 0.00901 atm<\/li>\n\n\n\n
      • Temperature = 22.5\u00b0C = 295.65 K<\/li>\n\n\n\n
      • R = 0.0821<\/li>\n<\/ul>\n\n\n\n

        Using PV = nRT: n=PVRT=(0.00901)(0.165)(0.0821)(295.65)\u22486.13\u00d710\u22126 moln = \\frac{PV}{RT} = \\frac{(0.00901)(0.165)}{(0.0821)(295.65)} \\approx 6.13 \\times 10^{-6}\\ \\text{mol}<\/p>\n\n\n\n

        Now calculate molar mass: M=massmol=0.01256.13\u00d710\u22126\u2248204 g\/molM = \\frac{\\text{mass}}{\\text{mol}} = \\frac{0.0125}{6.13 \\times 10^{-6}} \\approx 204\\ \\text{g\/mol}<\/p>\n\n\n\n

        Empirical formula = CHF\u2082
        Empirical mass = 12.01 (C) + 1.008 (H) + 2\u00d718.998 (F) \u2248 52.01 g\/mol<\/strong> Ratio=20452.01\u22483.92\u22484\\text{Ratio} = \\frac{204}{52.01} \\approx 3.92 \\approx 4<\/p>\n\n\n\n

        Molecular formula = (CHF2)4=C4H4F8(CHF\u2082)_4 = \\boxed{C\u2084H\u2084F\u2088}<\/p>\n\n\n\n

        \n\n\n\n

        \u2705 Final Answers:<\/h3>\n\n\n\n

        (a)<\/strong> 116.8 g KO\u2082
        (b)<\/strong> 1.46 atm O\u2082
        (c)<\/strong> Molecular formula: C\u2084H\u2084F\u2088<\/strong><\/p>\n\n\n\n

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

        (a)What mass of KO2, in grams, is required to react with 15.9 L of CO2 at 23.0\u00b0C and 955 mm Hg? _ g (b)(c) A self-contained underwater breathing apparatus uses canisters containing potassium superoxide. The superoxide consumes the exhaled CO2 by a person and replaces it with oxygen. 4KO2 (s) + 2CO2 (g) —> 2K2CO3 […]<\/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-218510","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218510","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=218510"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218510\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218510"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218510"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218510"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}