6 mL of 1.0M CuSO4(ag) is mixed with 8 mL of 1.0 M Na2S(ag) to produce 0.250g CuS, a black precipitate. The balanced chemical equation for this reaction is: CuSO4+Na2S —-> CuS + Na2SO4.

Question

a.) What is the limiting reactant in this reaction?

b.) Calculate the theoretical (maximum yield) of CuS in grams.

c.) What is the percent yield of this reaction?

Answer 1

CuSO₄ + Na₂S ==> CuS + Na₂SO₄ ... balanced equation

(a) One easy way to find the limiting reactant in a reaction is to divide the moles of each reactant by the corresponding coefficient in the balanced equation. In the current problem, 1 is the coefficient for both CuSO4 and for Na2S, so we simply divide the moles of each by one, and see which is less:

mols CuSO4 = 6 ml x 1 L/ 1000 ml x 1.0 mol/L = 0.006 mols CuSO4

mols Na2S = 8 ml x 1 L / 1000 ml x 1.0 mol/L = 0.008 mols Na2S

CuSO4 is the limiting reactant

(b) To calculate theoretical yield, we use the limiting reactant to determine how much product can be made:

0.006 CuSO4 x 1 mol CuS / mol CuSO4 x 95.6 g /mol CuS = 0.57 g CuS (2 sig. figs.)

(c) To find % yield, we divide the actual yield (0.250 g) by the theoretical yield and convert to %:

0.250 g / 0.57 g (x100) = 44% (2 sig. figs.)