An aqueous solution of ammonium sulfide is mixed with an aqueous solution of iron (III) Chloride

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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) chloride arc mixed, how many grams of precipitate will form?

c. What are the concentrations of ammonium ion and iron (III) ion left in solution after the reaction is complete?

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?

The Correct Answer and Explanation is :

Part A: Writing the Equations

1. Balanced Molecular Equation

FeCl3(aq)+(NH4)2S(aq)→Fe2S3(s)+NH4Cl(aq)\text{FeCl}_3 (aq) + (NH_4)_2S (aq) \rightarrow Fe_2S_3 (s) + NH_4Cl (aq)

Balanced equation: 2FeCl3(aq)+3(NH4)2S(aq)→Fe2S3(s)+6NH4Cl(aq)2FeCl_3 (aq) + 3(NH_4)_2S (aq) \rightarrow Fe_2S_3 (s) + 6NH_4Cl (aq)

2. Complete Ionic Equation

2Fe3+(aq)+6Cl−(aq)+6NH4+(aq)+3S2−(aq)→Fe2S3(s)+6NH4+(aq)+6Cl−(aq)2Fe^{3+} (aq) + 6Cl^- (aq) + 6NH_4^+ (aq) + 3S^{2-} (aq) \rightarrow Fe_2S_3 (s) + 6NH_4^+ (aq) + 6Cl^- (aq)

3. Net Ionic Equation

2Fe3+(aq)+3S2−(aq)→Fe2S3(s)2Fe^{3+} (aq) + 3S^{2-} (aq) \rightarrow Fe_2S_3 (s)

Part B: Mass of Precipitate Formed

Given:

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

Using stoichiometry:

  1. Moles of Fe³⁺ available: Moles=C×V=(0.250)×(0.1000)=0.0250 moles Fe3+\text{Moles} = C \times V = (0.250) \times (0.1000) = 0.0250 \text{ moles Fe}^{3+}
  2. Moles of S²⁻ available: Moles=C×V=(0.5X)×(0.0500)=0.025X moles S2−\text{Moles} = C \times V = (0.5X) \times (0.0500) = 0.025X \text{ moles S}^{2-}

From the balanced equation: 2Fe3++3S2−→Fe2S32Fe^{3+} + 3S^{2-} \rightarrow Fe_2S_3

  • Limiting reactant analysis:
    The required ratio is 2:3, meaning we need 1.5 times more sulfide than Fe³⁺.
    • If X<1.5X < 1.5, S²⁻ is limiting
    • If X>1.5X > 1.5, Fe³⁺ is limiting
    • If X=1.5X = 1.5, both react completely

For mass calculation, assuming Fe³⁺ is limiting (X>1.5X > 1.5):

  • Moles of Fe₂S₃ formed: 0.0250 moles Fe3+2=0.0125 moles Fe2S3\frac{0.0250 \text{ moles Fe}^{3+}}{2} = 0.0125 \text{ moles Fe}_2S_3
  • Molar mass of Fe₂S₃: 2(55.85)+3(32.06)=207.88 g/mol2(55.85) + 3(32.06) = 207.88 \text{ g/mol}
  • Mass of precipitate: 0.0125×207.88=2.60 g Fe2S30.0125 \times 207.88 = 2.60 \text{ g Fe}_2S_3

Part C: Remaining Ion Concentrations

  1. Ammonium ion (NH4+NH_4^+) concentration:
    • From (NH4)2S(NH_4)_2S, initial moles: 2(0.025X)=0.050X moles2(0.025X) = 0.050X \text{ moles}
    • Final volume: 50.0 mL+100.0 mL=150.0 mL=0.1500 L50.0 \text{ mL} + 100.0 \text{ mL} = 150.0 \text{ mL} = 0.1500 \text{ L}
    • Concentration: C=0.050X0.1500=X3 MC = \frac{0.050X}{0.1500} = \frac{X}{3} \text{ M}
  2. Iron(III) ion (Fe3+Fe^{3+}) concentration:
    • Since Fe³⁺ is the limiting reactant, all is consumed → [Fe³⁺] = 0 M

Part D: Preparing 100.0 mL of 0.250 M FeCl₃ Solution

To prepare 100.0 mL of 0.250 M FeCl₃, follow these steps:

  1. Calculate the required mass of FeCl₃:
    • Molar mass of FeCl₃: 55.85+3(35.45)=162.20 g/mol55.85 + 3(35.45) = 162.20 \text{ g/mol}
    • Moles needed: 0.250 M×0.1000 L=0.0250 moles FeCl30.250 \text{ M} \times 0.1000 \text{ L} = 0.0250 \text{ moles FeCl}_3
    • Mass of FeCl₃ required: 0.0250×162.20=4.055 g FeCl30.0250 \times 162.20 = 4.055 \text{ g FeCl}_3
  2. Procedure:
    • Weigh 4.06 g (rounding to 2 decimal places) of solid FeCl₃ using an analytical balance.
    • Transfer the solid carefully into a 100.0 mL volumetric flask.
    • Add about 50 mL of distilled water to dissolve the FeCl₃ while gently swirling.
    • Once dissolved, fill up to the 100.0 mL mark with distilled water.
    • Mix well by inverting the flask several times.

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