2Al + 6HCl -> 2AlCl3 + 3H2 (1)
Mg + 2HCl -> MgCl2 + H2 (2)
mH2 bay ra=3,9-3,5=0,4(g)
nH2=0,2(mol)
Đặt nAl=a \(\Leftrightarrow\)mAl=27a
nMg=b \(\Leftrightarrow\)mMg=24b
Ta có hệ:
\(\left\{{}\begin{matrix}27a+24b=3,9\\1,5a+b=0,2\end{matrix}\right.\)
=> a=0,1;b=0,05
mAl=27.0,1=2,7(g)
mMg=24.0,05=1,2(g)