Answer to Question #100282 in Organic Chemistry for Chelsey Morel

Question

14. 6 mL of 1.0M CuSO4(aq) is mixed with 8 mL of 1.0 M Na2S(aq) to produce 0.250 g CuS, a black precipitate. The balanced chemical equation for this reaction is: CuSO4 + Na2S  CuS + Na2SO4.
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

n = c•V

n(CuSO₄) = 1,0 mol/l • 0,0146 l = 0,0146 mol

n(Na₂S) = 1,0 mol/l • 0,008 l = 0,008 mol

So, answer a) is Na₂S.

n(Na₂S) = n(CuS)

m(CuS) = n•M = 0,008 mol • 95,61 g/mol = 0,765 g

Answer b) is 0,765 g

0,25/0,765 • 100% = 32,7 %

Answer c) is 32,7%