mCuSO4= 400. 20%= 80(g)
=> nCuSO4= 80/160= 0,5(mol)
PTHH: Fe + CuSO4-> FeSO4 + Cu
0,5_______0,5___0,5________0,5(mol)
mFe(phản ứng)= 0,5.56=28(g)
mCu(sản phẩm)= 0,5.64= 32(g)
m(rắn)= mCu(sản phẩm) + mFe2O3
<=> 80= 32+mFe2O3
=> mFe2O3= 48(g)
%mFe= \(\frac{28}{28+48}.100\approx38,642\%\)
=> %mFe2O3 \(\approx\) 100% - 36,842% \(\approx\) 63,158%
b) mddC= mFe + mddCuSO4 - mCu = 28 + 400 - 32= 396(g)
mFeSO4= 0,5.152= 76(g)
=> \(C\%ddFeSO4=\frac{76}{396}.100\approx19,192\%\)