Asked by Yaneth
CuSO4+4NH3-Cu(NH3)4SO4
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?
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?
Answers
Answered by
DrBob222
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 = ?.
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 = ?.
Answered by
rushi
the answer of DrBob222 is almost all correct but the theoretical yield=mol*molar mass
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