{"id":259162,"date":"2025-07-18T18:51:00","date_gmt":"2025-07-18T18:51:00","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=259162"},"modified":"2025-07-18T18:51:02","modified_gmt":"2025-07-18T18:51:02","slug":"sulfuric-acid-h2so4-reacts-with-sodium-bicarbonate-nahco3-to-produce-sodium-sulfate-carbon-dioxide-and-water-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/18\/sulfuric-acid-h2so4-reacts-with-sodium-bicarbonate-nahco3-to-produce-sodium-sulfate-carbon-dioxide-and-water-2\/","title":{"rendered":"Sulfuric acid (H2SO4) reacts with sodium bicarbonate (NaHCO3) to produce sodium sulfate, carbon dioxide, and water."},"content":{"rendered":"\n
Sulfuric acid (H2SO4) reacts with sodium bicarbonate (NaHCO3) to produce sodium sulfate, carbon dioxide, and water. If 2.48 L of 3.00 M H2SO4 is spilled and 1.25 kg NaHCO3 (MM=84.01 g\/mol) is available to clean up the spill, is there enough sodium bicarbonate to neutralize the spilled sulfuric acid? a. Yes, and there is about 20% more NaHCO3 than needed. b. Yes, and there is precisely enough NaHCO3. c. Yes, and there is about 10% more NaHCO3 than needed. d. No, and there is about 20% less NaHCO3 than needed. e. No, and there is about 10% less NaHCO3 than needed. In the reaction: 3H3AsO3(aq) + KBrO3(aq) \u00e2\u2020\u2019 KBr(aq) + 3H3AsO4(aq) Which element is oxidized? Br, H, O, As Which element is reduced? Br, H, O, As<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Sulfuric Acid (H2SO4) and Sodium Bicarbonate (NaHCO3) Neutralization<\/strong><\/h3>\n\n\n\n
The reaction between sulfuric acid and sodium bicarbonate can be written as: H2SO4(aq)+2NaHCO3(aq)\u2192Na2SO4(aq)+2CO2(g)+2H2O(l)\\text{H}_2\\text{SO}_4 (aq) + 2\\text{NaHCO}_3 (aq) \\rightarrow \\text{Na}_2\\text{SO}_4 (aq) + 2\\text{CO}_2 (g) + 2\\text{H}_2\\text{O} (l)H2\u200bSO4\u200b(aq)+2NaHCO3\u200b(aq)\u2192Na2\u200bSO4\u200b(aq)+2CO2\u200b(g)+2H2\u200bO(l)<\/p>\n\n\n\n
Step 1: Calculate moles of H2SO4<\/h4>\n\n\n\n
We are given:<\/p>\n\n\n\n
\n
Volume of H2SO4 = 2.48 L<\/li>\n\n\n\n
Molarity of H2SO4 = 3.00 M<\/li>\n<\/ul>\n\n\n\n
The number of moles of H2SO4 is calculated using the formula: moles of H2SO4=Volume (L)\u00d7Molarity (M)\\text{moles of H}_2\\text{SO}_4 = \\text{Volume (L)} \\times \\text{Molarity (M)}moles of H2\u200bSO4\u200b=Volume (L)\u00d7Molarity (M) moles of H2SO4=2.48\u2009L\u00d73.00\u2009mol\/L=7.44\u2009mol\\text{moles of H}_2\\text{SO}_4 = 2.48 \\, \\text{L} \\times 3.00 \\, \\text{mol\/L} = 7.44 \\, \\text{mol}moles of H2\u200bSO4\u200b=2.48L\u00d73.00mol\/L=7.44mol<\/p>\n\n\n\n
Step 2: Calculate moles of NaHCO3 required<\/h4>\n\n\n\n
From the balanced chemical equation, the stoichiometric ratio of NaHCO3 to H2SO4 is 2:1. Therefore, the moles of NaHCO3 needed to neutralize 7.44 moles of H2SO4 are: moles of NaHCO3=2\u00d77.44=14.88\u2009mol\\text{moles of NaHCO}_3 = 2 \\times 7.44 = 14.88 \\, \\text{mol}moles of NaHCO3\u200b=2\u00d77.44=14.88mol<\/p>\n\n\n\n
Step 3: Calculate moles of NaHCO3 available<\/h4>\n\n\n\n
We are given:<\/p>\n\n\n\n
\n
Mass of NaHCO3 = 1.25 kg = 1250 g<\/li>\n\n\n\n
Molar mass of NaHCO3 = 84.01 g\/mol<\/li>\n<\/ul>\n\n\n\n
The moles of NaHCO3 available is: moles of NaHCO3=Mass (g)Molar mass (g\/mol)\\text{moles of NaHCO}_3 = \\frac{\\text{Mass (g)}}{\\text{Molar mass (g\/mol)}}moles of NaHCO3\u200b=Molar mass (g\/mol)Mass (g)\u200b moles of NaHCO3=1250\u2009g84.01\u2009g\/mol=14.87\u2009mol\\text{moles of NaHCO}_3 = \\frac{1250 \\, \\text{g}}{84.01 \\, \\text{g\/mol}} = 14.87 \\, \\text{mol}moles of NaHCO3\u200b=84.01g\/mol1250g\u200b=14.87mol<\/p>\n\n\n\n
Step 4: Comparison<\/h4>\n\n\n\n
\n
Moles of NaHCO3 required = 14.88 mol<\/li>\n\n\n\n
Moles of NaHCO3 available = 14.87 mol<\/li>\n<\/ul>\n\n\n\n
We can see that the available NaHCO3 is slightly less than required (about 0.01 mol less), which is about 0.07% less than the required amount.<\/p>\n\n\n\n
Thus, the answer is e. No, and there is about 10% less NaHCO3 than needed<\/strong>, because the provided NaHCO3 would not be sufficient for the reaction.<\/p>\n\n\n\n\n\n\n\n
Oxidation and Reduction in the Reaction:<\/strong><\/h3>\n\n\n\n
The given reaction is: 3H3AsO3(aq)+KBrO3(aq)\u2192KBr(aq)+3H3AsO4(aq)3\\text{H}_3\\text{AsO}_3 (aq) + \\text{KBrO}_3 (aq) \\rightarrow \\text{KBr} (aq) + 3\\text{H}_3\\text{AsO}_4 (aq)3H3\u200bAsO3\u200b(aq)+KBrO3\u200b(aq)\u2192KBr(aq)+3H3\u200bAsO4\u200b(aq)<\/p>\n\n\n\n
Oxidation:<\/h4>\n\n\n\n
\n
Oxidation involves the loss of electrons.<\/li>\n\n\n\n
The element that undergoes oxidation will show an increase in its oxidation state.<\/li>\n<\/ul>\n\n\n\n
In this reaction, the Arsenic (As)<\/strong> in H\u2083AsO\u2083<\/strong> is oxidized. In H\u2083AsO\u2083, As has an oxidation state of +3, while in H\u2083AsO\u2084, As has an oxidation state of +5. Therefore, As is oxidized<\/strong>.<\/p>\n\n\n\n
Reduction:<\/h4>\n\n\n\n
\n
Reduction involves the gain of electrons.<\/li>\n\n\n\n
The element that undergoes reduction will show a decrease in its oxidation state.<\/li>\n<\/ul>\n\n\n\n
In this case, Bromine (Br)<\/strong> in KBrO\u2083<\/strong> is reduced. In KBrO\u2083, Br has an oxidation state of +5, while in KBr, Br has an oxidation state of -1. Therefore, Br is reduced<\/strong>.<\/p>\n\n\n\n
Sulfuric acid (H2SO4) reacts with sodium bicarbonate (NaHCO3) to produce sodium sulfate, carbon dioxide, and water. If 2.48 L of 3.00 M H2SO4 is spilled and 1.25 kg NaHCO3 (MM=84.01 g\/mol) is available to clean up the spill, is there enough sodium bicarbonate to neutralize the spilled sulfuric acid? a. Yes, and there is about […]<\/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-259162","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/259162","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=259162"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/259162\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=259162"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=259162"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=259162"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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