{"id":238591,"date":"2025-06-18T08:48:08","date_gmt":"2025-06-18T08:48:08","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=238591"},"modified":"2025-06-18T08:48:10","modified_gmt":"2025-06-18T08:48:10","slug":"2-00-g-of-strontium-bicarbonate-srhco32-reacts-with-250-ml-of-6-00-m-hcl-to-produce-carbon-dioxide-water-and-strontium-chloride","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/18\/2-00-g-of-strontium-bicarbonate-srhco32-reacts-with-250-ml-of-6-00-m-hcl-to-produce-carbon-dioxide-water-and-strontium-chloride\/","title":{"rendered":"2.00 g of strontium bicarbonate [Sr(HCO3)2] reacts with 250 mL of 6.00 M HCl to produce carbon dioxide, water, and strontium chloride."},"content":{"rendered":"\n

2.00 g of strontium bicarbonate [Sr(HCO3)2] reacts with 250 mL of 6.00 M HCl to produce carbon dioxide, water, and strontium chloride. Assuming the reaction proceeds to completion and the volume of solution remains constant: What is the final concentration of chloride ions in solution? What is the final concentration of strontium ions in solution? Note: Strontium chloride is soluble in water. Sr(HCO3)2 + 2 HCl —–> 2 CO2 + 2 H2O + SrCl2<\/p>\n\n\n\n

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

To solve this problem, we will follow these steps:<\/p>\n\n\n\n

\n\n\n\n

Step 1: Write the balanced chemical equation<\/strong><\/h3>\n\n\n\n

Sr(HCO3)2+2HCl\u21922CO2+2H2O+SrCl2\\text{Sr(HCO}_3\\text{)}_2 + 2\\text{HCl} \\rightarrow 2\\text{CO}_2 + 2\\text{H}_2\\text{O} + \\text{SrCl}_2Sr(HCO3\u200b)2\u200b+2HCl\u21922CO2\u200b+2H2\u200bO+SrCl2\u200b<\/p>\n\n\n\n

From the equation, 1 mole of strontium bicarbonate reacts with 2 moles of HCl to form 1 mole of SrCl\u2082<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Step 2: Determine the moles of each reactant<\/strong><\/h3>\n\n\n\n

Molar mass of Sr(HCO\u2083)\u2082:<\/strong><\/p>\n\n\n\n

    \n
  • Sr = 87.62 g\/mol<\/li>\n\n\n\n
  • H = 1.01 \u00d7 2 = 2.02 g\/mol<\/li>\n\n\n\n
  • C = 12.01 \u00d7 2 = 24.02 g\/mol<\/li>\n\n\n\n
  • O\u2083 = 16.00 \u00d7 6 = 96.00 g\/mol<\/li>\n\n\n\n
  • Total = 87.62 + 2.02 + 24.02 + 96.00 = 209.66 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n

    Moles of Sr(HCO\u2083)\u2082:<\/strong>2.00 g209.66 g\/mol\u22480.00954 mol\\frac{2.00\\text{ g}}{209.66\\text{ g\/mol}} \\approx 0.00954\\text{ mol}209.66 g\/mol2.00 g\u200b\u22480.00954 mol<\/p>\n\n\n\n

    Moles of HCl:<\/strong>Molarity\u00d7Volume=6.00 mol\/L\u00d70.250 L=1.50 mol\\text{Molarity} \\times \\text{Volume} = 6.00\\text{ mol\/L} \\times 0.250\\text{ L} = 1.50\\text{ mol}Molarity\u00d7Volume=6.00 mol\/L\u00d70.250 L=1.50 mol<\/p>\n\n\n\n

    \n\n\n\n

    Step 3: Determine the limiting reagent<\/strong><\/h3>\n\n\n\n

    From the balanced equation, 1 mole of Sr(HCO\u2083)\u2082 requires 2 moles of HCl.<\/p>\n\n\n\n

      \n
    • Required HCl = 0.00954 mol \u00d7 2 = 0.01908 mol<\/li>\n\n\n\n
    • Available HCl = 1.50 mol<\/li>\n\n\n\n
    • So HCl is in excess, and Sr(HCO\u2083)\u2082 is the limiting reagent<\/strong><\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Step 4: Calculate final concentrations<\/strong><\/h3>\n\n\n\n

      Moles of SrCl\u2082 produced = moles of Sr(HCO\u2083)\u2082 = 0.00954 mol<\/strong><\/p>\n\n\n\n

        \n
      • SrCl\u2082 dissociates in water: SrCl2\u2192Sr2++2Cl\u2212\\text{SrCl}_2 \\rightarrow \\text{Sr}^{2+} + 2\\text{Cl}^-SrCl2\u200b\u2192Sr2++2Cl\u2212<\/li>\n<\/ul>\n\n\n\n

        So:<\/p>\n\n\n\n

          \n
        • Moles of Sr\u00b2\u207a = 0.00954 mol<\/li>\n\n\n\n
        • Moles of Cl\u207b = 2 \u00d7 0.00954 mol = 0.01908 mol from SrCl\u2082<\/li>\n\n\n\n
        • Excess HCl (unreacted): 1.50\u00a0mol\u22120.01908\u00a0mol=1.48092\u00a0mol1.50\\text{ mol} – 0.01908\\text{ mol} = 1.48092\\text{ mol}1.50\u00a0mol\u22120.01908\u00a0mol=1.48092\u00a0mol<\/li>\n<\/ul>\n\n\n\n

          Total Cl\u207b ions = from SrCl\u2082 + from unreacted HCl
          = 0.01908 mol + 1.48092 mol = 1.50 mol<\/strong><\/p>\n\n\n\n

          Final concentrations (volume = 0.250 L):<\/strong><\/p>\n\n\n\n

            \n
          • [Cl\u207b] =<\/strong> 1.50\u00a0mol0.250\u00a0L=6.00\u00a0M\\frac{1.50\\text{ mol}}{0.250\\text{ L}} = \\boxed{6.00\\text{ M}}0.250\u00a0L1.50\u00a0mol\u200b=6.00\u00a0M\u200b<\/li>\n\n\n\n
          • [Sr\u00b2\u207a] =<\/strong> 0.00954\u00a0mol0.250\u00a0L=0.0382\u00a0M\\frac{0.00954\\text{ mol}}{0.250\\text{ L}} = \\boxed{0.0382\\text{ M}}0.250\u00a0L0.00954\u00a0mol\u200b=0.0382\u00a0M\u200b<\/li>\n<\/ul>\n\n\n\n
            \n\n\n\n

            Final Answers:<\/strong><\/h3>\n\n\n\n
              \n
            • Final concentration of chloride ions:<\/strong> 6.00 M<\/strong><\/li>\n\n\n\n
            • Final concentration of strontium ions:<\/strong> 0.0382 M<\/strong><\/li>\n<\/ul>\n\n\n\n
              \n\n\n\n

              Explanation<\/strong><\/h3>\n\n\n\n

              In this chemical reaction, strontium bicarbonate reacts with hydrochloric acid to form carbon dioxide, water, and strontium chloride. The stoichiometry of the reaction indicates a 1:2 molar ratio between strontium bicarbonate and HCl. That means one mole of Sr(HCO\u2083)\u2082 needs two moles of HCl to completely react and yield one mole of SrCl\u2082.<\/p>\n\n\n\n

              We begin by calculating the moles of Sr(HCO\u2083)\u2082 using its molar mass, which is approximately 209.66 g\/mol. A 2.00 g sample gives us about 0.00954 mol. Then, we calculate the moles of HCl, using the given concentration and volume, yielding 1.50 mol. Comparing these quantities shows that HCl is in excess, and strontium bicarbonate is the limiting reagent.<\/p>\n\n\n\n

              Next, we use the stoichiometric relationship to find how much SrCl\u2082 is produced. Since 0.00954 mol of Sr(HCO\u2083)\u2082 reacts fully, it produces the same amount of SrCl\u2082. SrCl\u2082 dissociates into Sr\u00b2\u207a and 2 Cl\u207b ions in solution. Additionally, the leftover HCl also contributes to the Cl\u207b concentration, since each HCl molecule provides one Cl\u207b ion.<\/p>\n\n\n\n

              Finally, dividing the total moles of each ion by the volume of solution (0.250 L) gives the final concentrations. Chloride ions come from both the SrCl\u2082 and the excess HCl, summing up to 1.50 mol, giving a final concentration of 6.00 M. The strontium ion concentration comes solely from the dissolved SrCl\u2082 and is 0.0382 M.<\/p>\n\n\n\n

              This problem demonstrates stoichiometry, limiting reagents, and solution concentration concepts.<\/p>\n\n\n\n

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

              2.00 g of strontium bicarbonate [Sr(HCO3)2] reacts with 250 mL of 6.00 M HCl to produce carbon dioxide, water, and strontium chloride. Assuming the reaction proceeds to completion and the volume of solution remains constant: What is the final concentration of chloride ions in solution? What is the final concentration of strontium ions in solution? […]<\/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-238591","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238591","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=238591"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238591\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=238591"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=238591"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=238591"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}