Answer to Question #324514 in General Chemistry for Ivan Ackerman

Question

Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.

SnBr4(aq)+2Na2S(aq)⟶4NaBr(aq)+SnS2(s)

Suppose a student adds 49.0 mL of a 0.452 M solution of SnBr4 to 44.8 mL of a 0.153 M  solution of Na2S.

Identify the limiting reactant.

Calculate the theoretical yield of SnS2

The student recovers 0.347 g SnS2. Calculate the percent yield of 

SnS2 that the student obtained.

.

Answer 1

SnBr₄(aq) + 2Na₂S(aq) ⟶ 4NaBr(aq) + SnS₂

(s)

n(SnBr₄) = C(sol) × V(sol) = 0.453 mol/L × 0.0490 L = 0.0222 mol

n(Na₂S) = 0.153 mol/L × 0.0448 L = 0.00685 mol

1 mol SnBr₄ reacts with 2 mol Na₂S

Na₂S is limiting reactant

2 mol of Na₂S produces 1 mol SnS₂

0.00685 mol Na₂S - x mol SnS₂

x = 0.00685 / 2 = 0.00343 mol

m(SnS₂) = n(SnS₂) × Mr(SnS₂) = 0.00343 mol × 182.81 g/mol = 0.627 g

percent yield = 0.347 g / 0.627 g × 100% = 55.34%