Gọi `x = n_(Al) ; y = n_(Fe) `
`n_(H_2) = 8.96 / 22.4 = 0.4 ( mol )`
`PTHH : 2Al + 3H_2SO_4 -> Al_2(SO_4)_3 + 3 H_2 ↑ `
` x - x/2 - 3/2x`
` Fe + H_2SO_4 -> FeSO_4 + H_2`
` y - y - y`
Ta có hệ pt : $\left \{ {{27x+56y=11} \atop {3/2x+y=0.4}} \right.$ `=>`$\left \{ {{x = 0.2} \atop {y=0.1}} \right.$
`-> %_(Al) =`$\dfrac{0.2×27}{11}$`×100% = 49.09%`
`-> %_(Fe) = 100% - 49.09% = 50.91%`
b)
`n_(Al_2(SO_4)_3) = 0.2 / 2 = 0.1 ( mol ) `
`-> m_(Al_2(SO_4)_3) = 0.1 × 342 = 34.2 ( g ) `
`n_(FeSO_4) = 0.1 mol `
`-> m_(FeSO_4) = 0.1 ×152 = 15.2 ( g ) `