CuSO4+4NH3-Cu(NH3)4SO4

Question

a. if you use 10.0g 0f CuSO4 and excess NH3, What is the theoretical yield of Cu(NH3)4SO4?

b. if you produce 12.6g of Cu(NH3)4SO4, What is the percent yield of the compund?

Answer 1

CuSO4+4NH3-Cu(NH3)4SO4

mols CuSO4 = g/molar mass.
Then ?mols CuSO4 = ? mol Cu(NH3)4SO4 where ? = the same number.
Then theoretical yield = grams x molar mass.

b.
%yield = (actual yield/theor yield)*100 = ?.

Answer 2

the answer of DrBob222 is almost all correct but the theoretical yield=mol*molar mass