How do I find the percent yield of Cu(NH3)4SO4xH2O if the theoretical yield of Cu(NH3)4SO4xH2O is...

Question

Question

How do I find the percent yield of Cu(NH3)4SO4xH2O if the theoretical yield of Cu(NH3)4SO4xH2O is 10.3g? the limiting reagent is 10g CuSO4x5H20

Answer 1

Answer #1

CuSO4 $\times$ 5H2O + 4NH3 ---------------> Cu(NH3)4SO4 $\times$ H2O + 4H2O

1 mole of CuSO4 $\times$ 5H2O to gives1mole of Cu(NH3)4SO4 $\times$ H2O

249.5g of CuSO4 $\times$ 5H2O to gives 245.5g of Cu(NH3)4SO4 $\times$ H2O

10g of CuSO4 $\times$ 5H2O to gives 245.5*10/249.5 = 9.84g of Cu(NH3)4SO4 $\times$ H2O

theoretical yield = 10.3g

percent yield = actual yield*100/theoretical yield

= 9.84*100/10.3 = 95.53% >>>>answer

Answer 2

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