Đáp án:
`a.`
`%m_{Fe}=50,909%`
`%m_{Al}=49,091%`
`b.219(g)`
Giải thích các bước giải:
`a.`
`Fe+2HCl->FeCl_2+H_2↑`
`2Al+6HCl->2AlCl_3+3H_2↑`
`n_{H_2}=(13,44)/(22,4)=0,6(mol)`
Gọi `a,b` là `n_{Fe},n_{Al}`
`=>m_{hh}=56a+27b=16,5(1)`
`∑n_{H_2}=a+3/2b=0,6(2)`
`(1),(2)=>a=0,15;b=0,3`
`%m_{Fe}=(0,15.56)/(16,5).100=50,909%`
`%m_{Al}=(0,3.27)/(16,5).100=49,091%`
`b.`
`n_{HCl}=2n_{H_2}=1,2(mol)`
`m_{HCl}=1,2.36,5=43,8(g)`
`m_{ddHCl}=43,8:20%=219(g)`