Mg + 2HCl -> MgCl2 + H2 (1)
2Al + 6HCl -> 2AlCl3 + 3H2 (2)
nH2=0.45(mol)
Đặt nMg=a
nAl=b
Ta có:
\(\left\{{}\begin{matrix}24a+27b=9\\a+1,5b=0,45\end{matrix}\right.\)
=>a=0,15;b=0,2
mMg=24.0,15=3,6(g)
%mMg=\(\dfrac{3,6}{9}.100\%=40\%\)
%mAl=100-40=60%