posted by lee on .
13. Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as
Cu2S(s) + O2(g) „_ 2Cu(s) + SO2(g)
If the reaction of 0.540 kg of copper(I) sulfide with excess oxygen produces 0.140 kg of
copper metal, what is the percent yield?
If that reaction were to go to completion, one mole of Cu2S(weighing 159 g) would form 2 moles of Cu (weighing 127g). 0.540 kg of Cu2S would thus lead to production of
0.540 * (127/159) = 0.431 g of Cu.
The 0.140 g actually produced is 32.5% of that amount. That is the percent yield.
Convert 0.540 kg Cu2S to moles. Moles = grams/molar mass.
Using the coefficients in the balanced equation, convert moles Cu2S to moles Cu metal.
Now convert moles Cu metal to grams. grams = moles x molar mass. This is the theoretical yield.
% yield = (0.140 kg/theoretical yield)*100 = ??
Be sure numerator and denominator are in the same units.