{"id":258987,"date":"2025-07-18T16:04:04","date_gmt":"2025-07-18T16:04:04","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=258987"},"modified":"2025-07-18T16:04:06","modified_gmt":"2025-07-18T16:04:06","slug":"when-baking-soda-reacts-with-sulfuric-acid-the-following-reaction-occurs-balance-the-chemical-equation-using-the-smallest-possible-whole-numbers","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/18\/when-baking-soda-reacts-with-sulfuric-acid-the-following-reaction-occurs-balance-the-chemical-equation-using-the-smallest-possible-whole-numbers\/","title":{"rendered":"When baking soda reacts with sulfuric acid, the following reaction occurs; balance the chemical equation using the smallest possible whole numbers"},"content":{"rendered":"\n

When baking soda reacts with sulfuric acid, the following reaction occurs; balance the chemical equation using the smallest possible whole numbers. Answer 1 Question 2 NaHCO3(s) + H2SO4(aq) \u2192 Na2SO4(aq) + Answer 2 Question 2 H2O(l) + Answer 3 Question 2 CO2(g) In an experiment, 22.4 mL of CO2 gas is collected over a 26.0 cm column of water at 298K, where the prevailing barometric pressure is 755.2 mm Hg. The vapor pressure of water at 298K is 23.8 Torr. Calculate a. Pressure of the water column: Answer 4 Question 2 Torr [three significant digits] b. Total pressure of the CO2 and water vapor in the gas collected: Answer 5 Question 2 Torr [three significant digits] c. Partial pressure of CO2 in the gas collected: Answer 6 Question 2 atm [three significant digits] d. Moles of CO2 collected: Answer 7 Question 2 mol [three significant digits] e. Moles of NaHCO3(s) that reacted: Answer 8 Question 2 mol [three significant digits] energy of a photon<\/p>\n\n\n\n

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

Balanced Chemical Equation:<\/h3>\n\n\n\n

The reaction between sodium bicarbonate (baking soda) and sulfuric acid is as follows:<\/p>\n\n\n\n

Unbalanced Equation:<\/strong> NaHCO3(s)+H2SO4(aq)\u2192Na2SO4(aq)+H2O(l)+CO2(g)\\text{NaHCO}_3(s) + \\text{H}_2\\text{SO}_4(aq) \\rightarrow \\text{Na}_2\\text{SO}_4(aq) + \\text{H}_2\\text{O}(l) + \\text{CO}_2(g)NaHCO3\u200b(s)+H2\u200bSO4\u200b(aq)\u2192Na2\u200bSO4\u200b(aq)+H2\u200bO(l)+CO2\u200b(g)<\/p>\n\n\n\n

To balance this equation, we ensure that the number of atoms for each element on both sides is the same.<\/p>\n\n\n\n

    \n
  1. Balance Na<\/strong> (Sodium): On the left, there is 1 Na from NaHCO\u2083. On the right, Na\u2082SO\u2084 has 2 Na, so we need 2 NaHCO\u2083.<\/li>\n\n\n\n
  2. Balance SO\u2084<\/strong> (Sulfate): On the left, there is 1 SO\u2084 from H\u2082SO\u2084. On the right, there is 1 SO\u2084 from Na\u2082SO\u2084, so it is already balanced.<\/li>\n\n\n\n
  3. Balance H<\/strong> (Hydrogen): On the left, there are 2 H from H\u2082SO\u2084. On the right, there are 2 H from H\u2082O, so it’s balanced.<\/li>\n\n\n\n
  4. Balance C<\/strong> (Carbon): On the left, there is 1 C from NaHCO\u2083. On the right, there is 1 C in CO\u2082, so it’s balanced.<\/li>\n\n\n\n
  5. Balance O<\/strong> (Oxygen): On the left, there are 3 O from NaHCO\u2083 and 4 O from H\u2082SO\u2084. On the right, there are 4 O from Na\u2082SO\u2084, 1 O from H\u2082O, and 2 O from CO\u2082. The oxygen atoms on both sides match.<\/li>\n<\/ol>\n\n\n\n

    Thus, the balanced equation is: 2NaHCO3(s)+H2SO4(aq)\u2192Na2SO4(aq)+2H2O(l)+2CO2(g)2 \\text{NaHCO}_3(s) + \\text{H}_2\\text{SO}_4(aq) \\rightarrow \\text{Na}_2\\text{SO}_4(aq) + 2 \\text{H}_2\\text{O}(l) + 2 \\text{CO}_2(g)2NaHCO3\u200b(s)+H2\u200bSO4\u200b(aq)\u2192Na2\u200bSO4\u200b(aq)+2H2\u200bO(l)+2CO2\u200b(g)<\/p>\n\n\n\n

    Gas Collection and Pressure Calculation:<\/h3>\n\n\n\n

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

      \n
    • Volume of CO\u2082: 22.4 mL = 0.0224 L<\/li>\n\n\n\n
    • Height of water column: 26.0 cm<\/li>\n\n\n\n
    • Temperature: 298K<\/li>\n\n\n\n
    • Barometric pressure: 755.2 mm Hg<\/li>\n\n\n\n
    • Vapor pressure of water at 298K: 23.8 Torr<\/li>\n<\/ul>\n\n\n\n

      (a) Pressure of the Water Column:<\/h4>\n\n\n\n

      The pressure exerted by the water column is equal to the height of the water column, in cm, converted to Torr (since 1 cm of water is approximately equal to 0.7355 Torr). Pressure of water column=26.0\u2009cm\u00d70.7355\u2009Torrcm=19.1\u2009Torr\\text{Pressure of water column} = 26.0 \\, \\text{cm} \\times 0.7355 \\, \\frac{\\text{Torr}}{\\text{cm}} = 19.1 \\, \\text{Torr}Pressure of water column=26.0cm\u00d70.7355cmTorr\u200b=19.1Torr<\/p>\n\n\n\n

      (b) Total Pressure of CO\u2082 and Water Vapor:<\/h4>\n\n\n\n

      The total pressure of the CO\u2082 and water vapor is the barometric pressure minus the vapor pressure of water (since the vapor pressure contributes to the total pressure). Total pressure=755.2\u2009mm Hg\u221223.8\u2009Torr=731.4\u2009Torr\\text{Total pressure} = 755.2 \\, \\text{mm Hg} – 23.8 \\, \\text{Torr} = 731.4 \\, \\text{Torr}Total pressure=755.2mm Hg\u221223.8Torr=731.4Torr<\/p>\n\n\n\n

      (c) Partial Pressure of CO\u2082 in the Gas Collected:<\/h4>\n\n\n\n

      The partial pressure of CO\u2082 is the total pressure minus the pressure exerted by the water vapor. Partial pressure of CO\u2082=731.4\u2009Torr\u221219.1\u2009Torr=712.3\u2009Torr\\text{Partial pressure of CO\u2082} = 731.4 \\, \\text{Torr} – 19.1 \\, \\text{Torr} = 712.3 \\, \\text{Torr}Partial pressure of CO\u2082=731.4Torr\u221219.1Torr=712.3Torr<\/p>\n\n\n\n

      Convert this to atmospheres: Partial pressure of CO\u2082=712.3760=0.937\u2009atm\\text{Partial pressure of CO\u2082} = \\frac{712.3}{760} = 0.937 \\, \\text{atm}Partial pressure of CO\u2082=760712.3\u200b=0.937atm<\/p>\n\n\n\n

      (d) Moles of CO\u2082 Collected:<\/h4>\n\n\n\n

      Use the ideal gas law to calculate the moles of CO\u2082. The ideal gas law is: PV=nRTPV = nRTPV=nRT<\/p>\n\n\n\n

      Where:<\/p>\n\n\n\n

        \n
      • PPP = 0.937 atm (partial pressure of CO\u2082)<\/li>\n\n\n\n
      • VVV = 0.0224 L (volume of CO\u2082)<\/li>\n\n\n\n
      • RRR = 0.0821 L\u00b7atm\/(mol\u00b7K) (ideal gas constant)<\/li>\n\n\n\n
      • TTT = 298 K (temperature)<\/li>\n<\/ul>\n\n\n\n

        Rearranging the ideal gas law to solve for nnn: n=PVRT=(0.937\u2009atm)(0.0224\u2009L)(0.0821\u2009L\\cdotpatm\/mol\\cdotpK)(298\u2009K)=0.001\u2009moln = \\frac{PV}{RT} = \\frac{(0.937 \\, \\text{atm})(0.0224 \\, \\text{L})}{(0.0821 \\, \\text{L\u00b7atm\/mol\u00b7K})(298 \\, \\text{K})} = 0.001 \\, \\text{mol}n=RTPV\u200b=(0.0821L\\cdotpatm\/mol\\cdotpK)(298K)(0.937atm)(0.0224L)\u200b=0.001mol<\/p>\n\n\n\n

        (e) Moles of NaHCO\u2083(s) that Reacted:<\/h4>\n\n\n\n

        From the balanced equation, we see that 2 moles of NaHCO\u2083 react with 2 moles of CO\u2082. So, the moles of NaHCO\u2083 that reacted is the same as the moles of CO\u2082 collected. Moles of NaHCO\u2083=0.001\u2009mol\\text{Moles of NaHCO\u2083} = 0.001 \\, \\text{mol}Moles of NaHCO\u2083=0.001mol<\/p>\n\n\n\n

        Energy of a Photon<\/h3>\n\n\n\n

        The energy of a photon is given by the equation: E=h\u03bdE = h \\nuE=h\u03bd<\/p>\n\n\n\n

        Where:<\/p>\n\n\n\n

          \n
        • EEE is the energy of the photon (in joules),<\/li>\n\n\n\n
        • hhh is Planck’s constant (6.626\u00d710\u2212346.626 \\times 10^{-34}6.626\u00d710\u221234 J\u00b7s),<\/li>\n\n\n\n
        • \u03bd\\nu\u03bd is the frequency of the photon (in Hz).<\/li>\n<\/ul>\n\n\n\n
          \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

          When baking soda reacts with sulfuric acid, the following reaction occurs; balance the chemical equation using the smallest possible whole numbers. Answer 1 Question 2 NaHCO3(s) + H2SO4(aq) \u2192 Na2SO4(aq) + Answer 2 Question 2 H2O(l) + Answer 3 Question 2 CO2(g) In an experiment, 22.4 mL of CO2 gas is collected over a 26.0 […]<\/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-258987","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258987","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=258987"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258987\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=258987"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=258987"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=258987"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}