Question

Na2S (aq) + Zn(NO3)2 (aq)= ZnS (s) + 2 NaNO3 (aq). Determine the mass of zinc nitrate which will be needed to make 216 g of zinc sulfide, with a yield of 73.2% _____ g.

Answer 1

Na2S (aq) + Zn(NO3)2 (aq) ----> ZnS (s) + 2 NaNO3 (aq)

Given the yield is 216 g

Percentage yield = 73.2 %

We know that % yield = ( actual yield/ theoretical yield) x100

                     73.2 = ( 216 / theoretical yield) x100

So Theoretical yield = 295 g

Molar mass of ZnS is 97.5 g/mol

Molar mass of Zn(NO3)2 is 189.4 g/mol

According to the balanced equation ,

1 mol=189.4 g of Zn(NO3)2 produces 1 mol = 97.5 g of ZnS

M g of Zn(NO3)2 produces 295 g of ZnS

M = ( 295x189.4) / 97.5

    = 573.2 g of zinc nitrate