Zn + H2SO4 -> ZnSO4 + H2 (1)
Fe + H2SO4 -> FeSO4 + H2 (2)
mZn;Fe=21,6-3=18,6(g)
nH2=0,3(mol)
Đặt nZn=a
nFe=b
Ta có hệ:
\(\left\{{}\begin{matrix}65a+56b=18,6\\a+b=0,3\end{matrix}\right.\)
=>a=0,2;b=0,1
mFe=56.0,1=5,6(g)
%mCu=\(\dfrac{3}{21,6}.100\%=13,9\%\)
%mFe=\(\dfrac{5,6}{21,6}.100\%=26\%\)
%mZn=100-13,9-26=60,1%