{"id":253118,"date":"2025-07-11T21:48:54","date_gmt":"2025-07-11T21:48:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=253118"},"modified":"2025-07-11T21:48:57","modified_gmt":"2025-07-11T21:48:57","slug":"hydrochloric-acid-hci-reacts-with-sodium-carbonate-na2co3-forming-sodium-chloride-nacl-water-h2o-and-carbon-dioxide-co3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/11\/hydrochloric-acid-hci-reacts-with-sodium-carbonate-na2co3-forming-sodium-chloride-nacl-water-h2o-and-carbon-dioxide-co3\/","title":{"rendered":"Hydrochloric acid (HCI) reacts with sodium carbonate (Na2CO3), forming sodium chloride (NaCl), water (H2O), and carbon dioxide (CO3)"},"content":{"rendered":"\n<p>Hydrochloric acid (HCI) reacts with sodium carbonate (Na<br>CO<br>), forming sodium chloride (NaCl), water (H<br>O), and carbon dioxide (CO<br>). This equation is balanced as written: 2HCl(aq) + Na<br>CO<br>(aq) ?2NaCl(aq) + H<br>O(l) + CO<br>(g) Part A What volume of 2.75 mol L<br>HCI in litres is needed to react completely (with nothing left over) with 0.500 L of 0.300 mol L<br>Na<br>CO<br>? Express your answer numerically in litres. View Available Hint(s) Submit Part B A 393-mL sample of unknown HCI solution reacts completely with Na<br>CO<br>to form 14.1 g CO<br>. What was the concentration of the HCl solution? Express the molar concentration numerically.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Let&#8217;s break this down into two parts:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Part A<\/h3>\n\n\n\n<p>We are given the following information:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The reaction is:<br>2HCl(aq)+Na2CO3(aq)\u21922NaCl(aq)+H2O(l)+CO2(g)2HCl(aq) + Na_2CO_3(aq) \\rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g)2HCl(aq)+Na2\u200bCO3\u200b(aq)\u21922NaCl(aq)+H2\u200bO(l)+CO2\u200b(g)<\/li>\n\n\n\n<li>Volume of sodium carbonate (Na\u2082CO\u2083) = <strong>0.500 L<\/strong><\/li>\n\n\n\n<li>Concentration of Na\u2082CO\u2083 = <strong>0.300 mol\/L<\/strong><\/li>\n\n\n\n<li>Concentration of HCl = <strong>2.75 mol\/L<\/strong><\/li>\n<\/ul>\n\n\n\n<p>We need to determine the volume of HCl required to react completely with Na\u2082CO\u2083.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Step 1: Calculate moles of Na\u2082CO\u2083<\/h4>\n\n\n\n<p>The number of moles of Na\u2082CO\u2083 is given by: moles&nbsp;of&nbsp;Na2CO3=concentration\u00d7volume=0.300\u2009mol\/L\u00d70.500\u2009L=0.150\u2009mol\\text{moles of Na}_2CO_3 = \\text{concentration} \\times \\text{volume} = 0.300 \\, \\text{mol\/L} \\times 0.500 \\, \\text{L} = 0.150 \\, \\text{mol}moles&nbsp;of&nbsp;Na2\u200bCO3\u200b=concentration\u00d7volume=0.300mol\/L\u00d70.500L=0.150mol<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Step 2: Use the stoichiometric ratio<\/h4>\n\n\n\n<p>From the balanced equation, we can see that 2 moles of HCl react with 1 mole of Na\u2082CO\u2083. This gives us the ratio: moles&nbsp;of&nbsp;HClmoles&nbsp;of&nbsp;Na2CO3=2so,moles&nbsp;of&nbsp;HCl=2\u00d70.150\u2009mol=0.300\u2009mol\\frac{\\text{moles of HCl}}{\\text{moles of Na}_2CO_3} = 2 \\quad \\text{so,} \\quad \\text{moles of HCl} = 2 \\times 0.150 \\, \\text{mol} = 0.300 \\, \\text{mol}moles&nbsp;of&nbsp;Na2\u200bCO3\u200bmoles&nbsp;of&nbsp;HCl\u200b=2so,moles&nbsp;of&nbsp;HCl=2\u00d70.150mol=0.300mol<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Step 3: Calculate volume of HCl<\/h4>\n\n\n\n<p>Now that we know the moles of HCl required, we can find the volume using the molarity equation: Volume&nbsp;of&nbsp;HCl=moles&nbsp;of&nbsp;HClconcentration&nbsp;of&nbsp;HCl=0.300\u2009mol2.75\u2009mol\/L=0.109\u2009L=109\u2009mL\\text{Volume of HCl} = \\frac{\\text{moles of HCl}}{\\text{concentration of HCl}} = \\frac{0.300 \\, \\text{mol}}{2.75 \\, \\text{mol\/L}} = 0.109 \\, \\text{L} = 109 \\, \\text{mL}Volume&nbsp;of&nbsp;HCl=concentration&nbsp;of&nbsp;HClmoles&nbsp;of&nbsp;HCl\u200b=2.75mol\/L0.300mol\u200b=0.109L=109mL<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Part B<\/h3>\n\n\n\n<p>We are given the following information:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of unknown HCl solution = <strong>393 mL<\/strong> = <strong>0.393 L<\/strong><\/li>\n\n\n\n<li>Mass of CO\u2082 produced = <strong>14.1 g<\/strong><\/li>\n\n\n\n<li>Molar mass of CO\u2082 = <strong>44.01 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Step 1: Calculate moles of CO\u2082 produced<\/h4>\n\n\n\n<p>First, we calculate the moles of CO\u2082 produced using the given mass: moles&nbsp;of&nbsp;CO2=mass&nbsp;of&nbsp;CO2molar&nbsp;mass&nbsp;of&nbsp;CO2=14.1\u2009g44.01\u2009g\/mol=0.3205\u2009mol\\text{moles of CO}_2 = \\frac{\\text{mass of CO}_2}{\\text{molar mass of CO}_2} = \\frac{14.1 \\, \\text{g}}{44.01 \\, \\text{g\/mol}} = 0.3205 \\, \\text{mol}moles&nbsp;of&nbsp;CO2\u200b=molar&nbsp;mass&nbsp;of&nbsp;CO2\u200bmass&nbsp;of&nbsp;CO2\u200b\u200b=44.01g\/mol14.1g\u200b=0.3205mol<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Step 2: Use the stoichiometric ratio<\/h4>\n\n\n\n<p>From the balanced equation, we see that 2 moles of HCl produce 1 mole of CO\u2082. Therefore, the moles of HCl required are: moles&nbsp;of&nbsp;HCl=2\u00d70.3205\u2009mol=0.641\u2009mol\\text{moles of HCl} = 2 \\times 0.3205 \\, \\text{mol} = 0.641 \\, \\text{mol}moles&nbsp;of&nbsp;HCl=2\u00d70.3205mol=0.641mol<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Step 3: Calculate the concentration of HCl<\/h4>\n\n\n\n<p>Now, we can find the molar concentration of the HCl solution: Concentration&nbsp;of&nbsp;HCl=moles&nbsp;of&nbsp;HClvolume&nbsp;of&nbsp;solution&nbsp;in&nbsp;L=0.641\u2009mol0.393\u2009L=1.63\u2009mol\/L\\text{Concentration of HCl} = \\frac{\\text{moles of HCl}}{\\text{volume of solution in L}} = \\frac{0.641 \\, \\text{mol}}{0.393 \\, \\text{L}} = 1.63 \\, \\text{mol\/L}Concentration&nbsp;of&nbsp;HCl=volume&nbsp;of&nbsp;solution&nbsp;in&nbsp;Lmoles&nbsp;of&nbsp;HCl\u200b=0.393L0.641mol\u200b=1.63mol\/L<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Part A:<\/strong> The volume of 2.75 mol\/L HCl required is <strong>109 mL<\/strong>.<\/li>\n\n\n\n<li><strong>Part B:<\/strong> The concentration of the unknown HCl solution is <strong>1.63 mol\/L<\/strong>.<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner9-77.jpeg\" alt=\"\" class=\"wp-image-253119\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Hydrochloric acid (HCI) reacts with sodium carbonate (NaCO), forming sodium chloride (NaCl), water (HO), and carbon dioxide (CO). This equation is balanced as written: 2HCl(aq) + NaCO(aq) ?2NaCl(aq) + HO(l) + CO(g) Part A What volume of 2.75 mol LHCI in litres is needed to react completely (with nothing left over) with 0.500 L of [&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-253118","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/253118","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=253118"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/253118\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=253118"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=253118"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=253118"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}