`2Al+6HCl->2AlCl_3+3H_2`
`Fe+2HCl->FeCl_2+H_2`
`n_{Al;Fe}=x;ymol`
`n_{H_2}=3/2x+y=(7,84)/(22,4)=0,35mol`
`m_{hh}=27x+56y=13,9`
`x=0,1mol;y=0,2mol`
`%m_{Al}=(0,1.27)/(13,9%)=19,42%`
`%m_{Fe}=100%-19,42%=80,58%`
`n_{HCl}=3.0,1+2.0,2=0,7mol`
Đề thiếu dữ kiện tính `mdd_{spu}`