Question

The metathesis reaction between sodium sulfide and chromium(IIII) chloride is shown below

3Na2S(aq)+2 CrCl3(aq) arrow Cr2S3(s)+6NaCl(aq)

How many grams of solid chromium (III) sulfide can be produced by the reaction of 250.0 ml of 0.500 M sodium sulfide solution with 275.0 ml of 0.250 M Chromium (III) chloride solution?

Answer 6.882

determine the final concentration of each ion in solution at the end of the precipitation reaction

Cr^3
S^2
Na^2
Cl^-

Answer 1

Initial,

Amount of Na₂S = M * V / 1000 = 0.500 * 250.0 / 1000 = 0.125 mol

Amount of CrCl₃ = 0.250 * 275.0 / 1000 = 0.06875 mol

From the balalnced equation,

3 mol Na₂S needs 2 mol CrCl₃

Then,

0.125 mol Na₂S needs 0.125 * 2 / 3 = 0.0833 mol of CrCl₃

But we have only 0.06875 mol. So, CrCl₃ is limiting reagent.

From the balanced equation,

2 mol of CrCl₃ forms 1 mol Cr₂S₃

then,

0.06875 mol of CrCl₃ forms 0.06875 * 1 / 2 = 0.03438 mol of Cr₂S₃

Mass of Cr₂S₃ = AMount * molar mass = 0.03438 * 200 = 6.876 g.

By the end of reaction,

[Cr³⁺] = 0.000 M because entire Cr³⁺ is in the form of preci[itate.

[Cl^-] = 6 * 0.03438 * 1000 / (250.0 + 275.0) = 0.3929 M

Remaining amount of Na2S = 0.125 - (3*0.03438/2) = 0.07343 mol

[S^2-] = 3 * 0.07343 * 1000 / (250.0 + 275.0) = 0.4196 M

[Na⁺] = 6 * 0.125 * 1000 / (250.0+275.0) = 1.428 M