{"id":236922,"date":"2025-06-16T16:13:05","date_gmt":"2025-06-16T16:13:05","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=236922"},"modified":"2025-06-16T16:13:07","modified_gmt":"2025-06-16T16:13:07","slug":"balanced-equation-nahco3-hcl-nacl-h2o-co2-find-the-moles-from-1-03-grams-of-co2-produced","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/16\/balanced-equation-nahco3-hcl-nacl-h2o-co2-find-the-moles-from-1-03-grams-of-co2-produced\/","title":{"rendered":"Balanced equation: NaHCO3 + HCl -> NaCl + H2O + CO2 Find the moles from 1.03 grams of CO2 produced."},"content":{"rendered":"\n


Balanced equation: NaHCO3 + HCl -> NaCl + H2O + CO2 Find the moles from 1.03 grams of CO2 produced.<\/p>\n\n\n\n

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

To find the number of moles<\/strong> of carbon dioxide (CO\u2082<\/strong>) produced from 1.03 grams<\/strong>, we use the molar mass<\/strong> of CO\u2082 and the formula:Moles=Mass (g)Molar Mass (g\/mol)\\text{Moles} = \\frac{\\text{Mass (g)}}{\\text{Molar Mass (g\/mol)}}Moles=Molar Mass (g\/mol)Mass (g)\u200b<\/p>\n\n\n\n

Step 1: Determine the molar mass of CO\u2082<\/h3>\n\n\n\n
    \n
  • Carbon (C): 12.01 g\/mol<\/li>\n\n\n\n
  • Oxygen (O): 16.00 g\/mol<\/li>\n\n\n\n
  • CO\u2082 has two oxygen atoms<\/li>\n<\/ul>\n\n\n\n

    So,Molar Mass of CO\u2082=12.01+(2\u00d716.00)=12.01+32.00=44.01\u2009g\/mol\\text{Molar Mass of CO\u2082} = 12.01 + (2 \\times 16.00) = 12.01 + 32.00 = 44.01 \\, \\text{g\/mol}Molar Mass of CO\u2082=12.01+(2\u00d716.00)=12.01+32.00=44.01g\/mol<\/p>\n\n\n\n

    Step 2: Use the formula to find moles<\/h3>\n\n\n\n

    Moles of CO\u2082=1.03\u2009g44.01\u2009g\/mol\u22480.0234\u2009mol\\text{Moles of CO\u2082} = \\frac{1.03 \\, \\text{g}}{44.01 \\, \\text{g\/mol}} \\approx 0.0234 \\, \\text{mol}Moles of CO\u2082=44.01g\/mol1.03g\u200b\u22480.0234mol<\/p>\n\n\n\n

    \n\n\n\n

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

    0.0234 moles of CO\u2082<\/strong><\/p>\n\n\n\n

    \n\n\n\n

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

    The reaction between sodium bicarbonate (NaHCO\u2083) and hydrochloric acid (HCl) results in the formation of three products: sodium chloride (NaCl), water (H\u2082O), and carbon dioxide (CO\u2082). The balanced chemical equation is:NaHCO\u2083+HCl\u2192NaCl+H\u2082O+CO\u2082\\text{NaHCO\u2083} + \\text{HCl} \\rightarrow \\text{NaCl} + \\text{H\u2082O} + \\text{CO\u2082}NaHCO\u2083+HCl\u2192NaCl+H\u2082O+CO\u2082<\/p>\n\n\n\n

    This equation tells us that one mole of sodium bicarbonate reacts with one mole of hydrochloric acid to produce one mole each of sodium chloride, water, and carbon dioxide. Therefore, if we know the amount of CO\u2082 produced, we can determine the amount of NaHCO\u2083 that reacted.<\/p>\n\n\n\n

    In this problem, we are given the mass of carbon dioxide: 1.03 grams<\/strong>. To find the number of moles of CO\u2082, we must divide the mass by its molar mass. The molar mass of CO\u2082 is obtained by summing the atomic masses of its elements. Carbon has an atomic mass of 12.01 grams per mole, and oxygen has an atomic mass of 16.00 grams per mole. Since there are two oxygen atoms in CO\u2082, the molar mass becomes:12.01+2\u00d716.00=44.01\u2009g\/mol12.01 + 2 \\times 16.00 = 44.01 \\, \\text{g\/mol}12.01+2\u00d716.00=44.01g\/mol<\/p>\n\n\n\n

    Now, using the formula for moles (mass divided by molar mass), we calculate:1.0344.01\u22480.0234\u2009mol\\frac{1.03}{44.01} \\approx 0.0234 \\, \\text{mol}44.011.03\u200b\u22480.0234mol<\/p>\n\n\n\n

    Thus, 1.03 grams of carbon dioxide<\/strong> is equivalent to 0.0234 moles<\/strong>, and this value reflects the amount of CO\u2082 produced in the reaction.<\/p>\n\n\n\n

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

    Balanced equation: NaHCO3 + HCl -> NaCl + H2O + CO2 Find the moles from 1.03 grams of CO2 produced. The Correct Answer and Explanation is: To find the number of moles of carbon dioxide (CO\u2082) produced from 1.03 grams, we use the molar mass of CO\u2082 and the formula:Moles=Mass (g)Molar Mass (g\/mol)\\text{Moles} = \\frac{\\text{Mass (g)}}{\\text{Molar Mass (g\/mol)}}Moles=Molar Mass (g\/mol)Mass (g)\u200b […]<\/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-236922","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/236922","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=236922"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/236922\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=236922"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=236922"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=236922"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}