`Fe+H_2SO_4->FeSO_4+H_2↑`
`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2↑`
`n_{H_2SO_4}=0,17.0,5=0,085(mol)`
Gọi `a,b` là `n_{Fe},n_{Al}`
`=>m_{hh}=56a+27b=3,05(1)`
`∑n_{H_2SO_4}=a+3/2b=0,085(2)`
`(1),(2)=>a=0,04;b=0,03`
`%m_{Fe}=(0,04.56)/(3,05).100≈73,443%`
`%m_{Al}=(0,03.27)/(3,05).100≈26,557%`
`\text{___________________________________}`
Đáp án:
`%m_{Fe}≈73,443%`
`%m_{Al}≈26,557%`