If I have 0.56g of Iron powder and 6g of copper II sulfate and the iron powder is in excess, what is my theoretical yield?
Please help, I've tried to get the answer like a million times and I just can't get it. :/
Thanks in advance.
Please help, I've tried to get the answer like a million times and I just can't get it. :/
Thanks in advance.
-
Fe(s) + CuSO4(aq) >>>>>>>>>>>>>>>>> FeSO4(aq) + Cu(s)
It depends on the theoretical yield of WHICH SUBSTANCE. Lucky for you, I'll calculate both.
theoretical yield FeSO4 = (6 grams CuSO4) * (1 mol CuSO4 / 159.62 grams CuSO4) * (1 mol FeSO4 / 1 mol CuSO4) * (151.908 grams FeSO4 / 1 mol FeSO4) = 5.71 grams FeSO4 is the answer.
theoretical yield Cu = (6 grams CuSO4) * (1 mol CuSO4 / 159.62 grams CuSO4) * (1 mol Cu / 1 mol CuSO4) * (63.546 grams Cu / 1 mol Cu) = 2.39 grams Cu is the other answer.
It depends on the theoretical yield of WHICH SUBSTANCE. Lucky for you, I'll calculate both.
theoretical yield FeSO4 = (6 grams CuSO4) * (1 mol CuSO4 / 159.62 grams CuSO4) * (1 mol FeSO4 / 1 mol CuSO4) * (151.908 grams FeSO4 / 1 mol FeSO4) = 5.71 grams FeSO4 is the answer.
theoretical yield Cu = (6 grams CuSO4) * (1 mol CuSO4 / 159.62 grams CuSO4) * (1 mol Cu / 1 mol CuSO4) * (63.546 grams Cu / 1 mol Cu) = 2.39 grams Cu is the other answer.
-
If 6 g Cu(II) SO4 is anhydrous then it is 0.0376 moles of Cu( II) while Fe is 0.01 mole So Fe is not in excess it is Cu which is in excess
Theoretical yield should be 0.01 mole of whatever you are preparing
Theoretical yield should be 0.01 mole of whatever you are preparing