2Al + 6HCl -> 2AlCl3 + 3H2 (1)
Zn + 2HCl -> ZnCl2 + H2 (2)
nH2=0,4(mol)
Đặt nAl=a
nZn=b
Ta có hệ:
\(\left\{{}\begin{matrix}27a+65b=11,9\\1,5a+b=0,4\end{matrix}\right.\)
=>a=0,2;b=0,1
%mAl=\(\dfrac{27.0,2}{11,9}.100\%=45,38\%\)
%mZn=100-45,38=54,62%