`a.`
`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2↑`
`Zn+H_2SO_4->ZnSO_4+H_2↑`
`n_{H_2}=(10,08)/(22,4)=0,45(mol)`
Gọi `a,b` là `n_{Al},n_{Zn}`
`=>m_{hh}=27a+65b=22,2(1)`
`∑n_{H_2}=3/2a+b=0,45(2)`
`(1),(2)=>a=0,1;b=0,3`
`%m_{Al}=(0,1.27)/(22,2).100≈12,162%`
`%m_{Zn}=(0,3.65)/(22,2).100≈87,838%`
`b.`
`n_{H_2SO_4}=n_{H_2}=0,45(mol)`
`m_{H_2SO_4}=0,45.98=44,1(g)`
`C%_{ddH_2SO_4}=(44,1)/100 .100=44,1%`
`\text{___________________________________}`
Đáp án:
`a.`
`%m_{Al}≈12,162%`
`%m_{Zn}≈87,838%`
`b.`
`C%_{ddH_2SO_4}=44,1%`