CuO+ H2SO4------------>CuSO4+ H2O
nCuO=0.02 mol
nH2SO4=\(\dfrac{11.76\cdot25\%}{98}\)=0.03 mol
Xét tỉ lệ nCuO/1<nH2SO4/1
=>CuO hết, H2SO4 dư tính theo CuO
Theo PTHH nH2SO4=nCuSO4=nCuO=0.02 mol
mdd=1.6+11.76=13.36(g)
Do đó %mH2SO4 dư=\(\dfrac{\left(0.03-0.02\right)\cdot98\cdot100}{13.36}\)=7.33%
%mCuSO4=\(\dfrac{0.02\cdot160\cdot100}{13.36}\)=23.95%