`Mg+2HCl->MgCl_2+H_2`
`Fe+2HCl->FeCl_2+H_2`
`2Al+6HCl->2AlCl_3+3H_2`
`Δ_(m)=7,8-m_(H_2)=7`
`=>m_(H_2)=0,8(g)`
`=>n_(H_2)=\frac{0,8}{2}=0,4(mol)`
Theo `PT`
`n_(HCl)=2 .n_(H_2)=0,8(mol)`
Áp dung `ĐLBTKL`
`m_(hh)+m_(HCl)=m+m_(H_2)`
`=>m=7,8+0,8.36,5-0,8=36,2(g)`