One of the reactions involved in the smelting of copper sulfide ores involves copper(I) oxide and copper (I) sulfide: 2Cu2O(solid) + Cu2S(solid) ---> 6Cu(solid) + SO2(gas) Assuming that 35.00 g of copper (I) oxide is heated with 25.00 g of copper (I) sulfide.

Question

a) Determine which reagent is present in excess
b) Calculate the theoretical yield of copper
c) Determine the percentage yield if 35.46 g of copper is actually isolated

Answer 1

mols Cu2O = grams/molar mass = ?
mols Cu2S = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols Cu2O to mols Cu.
Do the same for mols Cu2S to mols Cu.
It is likely these two values will not be the same; in limiting reagent problems the smaller number is ALWAYS the right value to choose and the reagent producing that value is the limiting reagent. The other reagent will be the one in excess.

Using the smaller value, convert mols to grams of the product. g = mols x atomic mass Cu. This is the theoretical yield (TY). The actual yield (AY) is 35.46 from the problem.
% yield = (AY/TY)*100 = ?