{"id":194486,"date":"2025-02-24T06:46:48","date_gmt":"2025-02-24T06:46:48","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=194486"},"modified":"2025-02-24T06:46:51","modified_gmt":"2025-02-24T06:46:51","slug":"an-aqueous-solution-of-ammonium-sulfide-is-mixed-with-an-aqueous-solution-of-iron-iii-chloride","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/24\/an-aqueous-solution-of-ammonium-sulfide-is-mixed-with-an-aqueous-solution-of-iron-iii-chloride\/","title":{"rendered":"An aqueous solution of ammonium sulfide is mixed with an aqueous solution of iron (III) Chloride"},"content":{"rendered":"\n

An aqueous solution of ammonium sulfide is mixed with an aqueous solution of iron (III) Chloride.<\/p>\n\n\n\n

a. Write a balanced molecular equation, complete ionic equation, and net ionic equation for the two solutions mixed above. Include all phases.<\/p>\n\n\n\n

b. If 50.0 mL of 0.5(X) M ammonium sulphide and 100.0 mL of 0.250 M iron (III) chloride arc mixed, how many grams of precipitate will form?<\/p>\n\n\n\n

c. What are the concentrations of ammonium ion and iron (III) ion left in solution after the reaction is complete?<\/p>\n\n\n\n

d. In part b, you started with 100.0 mL of 0.250 M iron chloride. How would you prepare such a solution in the lab if you started with solid iron(III) chloride?<\/p>\n\n\n\n

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

Part A: Writing the Equations<\/strong><\/h3>\n\n\n\n

1. Balanced Molecular Equation<\/strong><\/h4>\n\n\n\n

FeCl3(aq)+(NH4)2S(aq)\u2192Fe2S3(s)+NH4Cl(aq)\\text{FeCl}_3 (aq) + (NH_4)_2S (aq) \\rightarrow Fe_2S_3 (s) + NH_4Cl (aq)<\/p>\n\n\n\n

Balanced equation: 2FeCl3(aq)+3(NH4)2S(aq)\u2192Fe2S3(s)+6NH4Cl(aq)2FeCl_3 (aq) + 3(NH_4)_2S (aq) \\rightarrow Fe_2S_3 (s) + 6NH_4Cl (aq)<\/p>\n\n\n\n

2. Complete Ionic Equation<\/strong><\/h4>\n\n\n\n

2Fe3+(aq)+6Cl\u2212(aq)+6NH4+(aq)+3S2\u2212(aq)\u2192Fe2S3(s)+6NH4+(aq)+6Cl\u2212(aq)2Fe^{3+} (aq) + 6Cl^- (aq) + 6NH_4^+ (aq) + 3S^{2-} (aq) \\rightarrow Fe_2S_3 (s) + 6NH_4^+ (aq) + 6Cl^- (aq)<\/p>\n\n\n\n

3. Net Ionic Equation<\/strong><\/h4>\n\n\n\n

2Fe3+(aq)+3S2\u2212(aq)\u2192Fe2S3(s)2Fe^{3+} (aq) + 3S^{2-} (aq) \\rightarrow Fe_2S_3 (s)<\/p>\n\n\n\n

\n\n\n\n

Part B: Mass of Precipitate Formed<\/strong><\/h3>\n\n\n\n

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

    \n
  • V(NH4)2S=50.0V_{\\text{(NH}_4)_2S} = 50.0 mL = 0.0500 L<\/li>\n\n\n\n
  • C(NH4)2S=0.5XC_{\\text{(NH}_4)_2S} = 0.5X M<\/li>\n\n\n\n
  • VFeCl3=100.0V_{\\text{FeCl}_3} = 100.0 mL = 0.1000 L<\/li>\n\n\n\n
  • CFeCl3=0.250C_{\\text{FeCl}_3} = 0.250 M<\/li>\n<\/ul>\n\n\n\n

    Using stoichiometry:<\/p>\n\n\n\n

      \n
    1. Moles of Fe\u00b3\u207a available:<\/strong> Moles=C\u00d7V=(0.250)\u00d7(0.1000)=0.0250\u00a0moles\u00a0Fe3+\\text{Moles} = C \\times V = (0.250) \\times (0.1000) = 0.0250 \\text{ moles Fe}^{3+}<\/li>\n\n\n\n
    2. Moles of S\u00b2\u207b available:<\/strong> Moles=C\u00d7V=(0.5X)\u00d7(0.0500)=0.025X\u00a0moles\u00a0S2\u2212\\text{Moles} = C \\times V = (0.5X) \\times (0.0500) = 0.025X \\text{ moles S}^{2-}<\/li>\n<\/ol>\n\n\n\n

      From the balanced equation: 2Fe3++3S2\u2212\u2192Fe2S32Fe^{3+} + 3S^{2-} \\rightarrow Fe_2S_3<\/p>\n\n\n\n

        \n
      • Limiting reactant analysis:<\/strong>
        The required ratio is 2:3, meaning we need 1.5 times<\/strong> more sulfide than Fe\u00b3\u207a.\n
          \n
        • If X<1.5X < 1.5, S\u00b2\u207b is limiting<\/strong><\/li>\n\n\n\n
        • If X>1.5X > 1.5, Fe\u00b3\u207a is limiting<\/strong><\/li>\n\n\n\n
        • If X=1.5X = 1.5, both react completely<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

          For mass calculation, assuming Fe\u00b3\u207a is limiting (X>1.5X > 1.5):<\/p>\n\n\n\n

            \n
          • Moles of Fe\u2082S\u2083 formed:<\/strong> 0.0250\u00a0moles\u00a0Fe3+2=0.0125\u00a0moles\u00a0Fe2S3\\frac{0.0250 \\text{ moles Fe}^{3+}}{2} = 0.0125 \\text{ moles Fe}_2S_3<\/li>\n\n\n\n
          • Molar mass of Fe\u2082S\u2083:<\/strong> 2(55.85)+3(32.06)=207.88\u00a0g\/mol2(55.85) + 3(32.06) = 207.88 \\text{ g\/mol}<\/li>\n\n\n\n
          • Mass of precipitate:<\/strong> 0.0125\u00d7207.88=2.60\u00a0g\u00a0Fe2S30.0125 \\times 207.88 = 2.60 \\text{ g Fe}_2S_3<\/li>\n<\/ul>\n\n\n\n
            \n\n\n\n

            Part C: Remaining Ion Concentrations<\/strong><\/h3>\n\n\n\n
              \n
            1. Ammonium ion (NH4+NH_4^+) concentration:<\/strong>\n
                \n
              • From (NH4)2S(NH_4)_2S, initial moles: 2(0.025X)=0.050X\u00a0moles2(0.025X) = 0.050X \\text{ moles}<\/li>\n\n\n\n
              • Final volume: 50.0\u00a0mL+100.0\u00a0mL=150.0\u00a0mL=0.1500\u00a0L50.0 \\text{ mL} + 100.0 \\text{ mL} = 150.0 \\text{ mL} = 0.1500 \\text{ L}<\/li>\n\n\n\n
              • Concentration: C=0.050X0.1500=X3\u00a0MC = \\frac{0.050X}{0.1500} = \\frac{X}{3} \\text{ M}<\/li>\n<\/ul>\n<\/li>\n\n\n\n
              • Iron(III) ion (Fe3+Fe^{3+}) concentration:<\/strong>\n
                  \n
                • Since Fe\u00b3\u207a is the limiting reactant, all is consumed<\/strong> \u2192 [Fe\u00b3\u207a] = 0 M<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n
                  \n\n\n\n

                  Part D: Preparing 100.0 mL of 0.250 M FeCl\u2083 Solution<\/strong><\/h3>\n\n\n\n

                  To prepare 100.0 mL of 0.250 M FeCl\u2083<\/strong>, follow these steps:<\/p>\n\n\n\n

                    \n
                  1. Calculate the required mass of FeCl\u2083:<\/strong>\n
                      \n
                    • Molar mass of FeCl\u2083: 55.85+3(35.45)=162.20\u00a0g\/mol55.85 + 3(35.45) = 162.20 \\text{ g\/mol}<\/li>\n\n\n\n
                    • Moles needed: 0.250\u00a0M\u00d70.1000\u00a0L=0.0250\u00a0moles\u00a0FeCl30.250 \\text{ M} \\times 0.1000 \\text{ L} = 0.0250 \\text{ moles FeCl}_3<\/li>\n\n\n\n
                    • Mass of FeCl\u2083 required: 0.0250\u00d7162.20=4.055\u00a0g\u00a0FeCl30.0250 \\times 162.20 = 4.055 \\text{ g FeCl}_3<\/li>\n<\/ul>\n<\/li>\n\n\n\n
                    • Procedure:<\/strong>\n
                        \n
                      • Weigh 4.06 g<\/strong> (rounding to 2 decimal places) of solid FeCl\u2083<\/strong> using an analytical balance.<\/li>\n\n\n\n
                      • Transfer the solid carefully<\/strong> into a 100.0 mL<\/strong> volumetric flask.<\/li>\n\n\n\n
                      • Add about 50 mL<\/strong> of distilled water<\/strong> to dissolve the FeCl\u2083 while gently swirling.<\/li>\n\n\n\n
                      • Once dissolved, fill up to the 100.0 mL mark<\/strong> with distilled water.<\/li>\n\n\n\n
                      • Mix well by inverting the flask several times.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n

                        Generated Image<\/strong><\/h3>\n\n\n\n

                        I\u2019ll generate an image illustrating the preparation of the FeCl\u2083 solution.<\/p>\n\n\n\n

                        Here is the generated image illustrating the preparation of the 100.0 mL 0.250 M FeCl\u2083 solution. Let me know if you need any modifications!<\/p>\n\n\n\n

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

                        An aqueous solution of ammonium sulfide is mixed with an aqueous solution of iron (III) Chloride. a. Write a balanced molecular equation, complete ionic equation, and net ionic equation for the two solutions mixed above. Include all phases. b. If 50.0 mL of 0.5(X) M ammonium sulphide and 100.0 mL of 0.250 M iron (III) […]<\/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-194486","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194486","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=194486"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194486\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=194486"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=194486"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=194486"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}