$\rm \text{Đặt: } \begin{cases} n_{Al} = x\ (mol)\\n_{Fe} = y\ (mol)` \end{cases}$

`2Al+6HCl->2AlCl_3+3H_2`

`Fe+2HCl->FeCl_2+H_2`

`-> 27x+56y = 11\ (1)`

`Theo\ pt:1,5n_{Al} + n_{Fe} = (8,96)/(22,4) = 0,4\ (2)`

`(1)(2)->x = 0,2; y = 0,1`

`a,`

`-> m_{Al} = 0,2 xx 27 = 5,4\ (g)`

`-> m_{Fe} = 11 - 5,4 = 5,6\ (g)`

`b,`

`Theo\ p:n_{HCl} = 3n_{Al} + 2n_{Fe} = 3 xx 0,2 + 2 xx 0,1 = 0,8\ (mol)`

`-> C\%_{HCl} = (0,8 xx 36,5)/(500) xx 100 = 5,84\%`

`a)`

`Hh:Al:x(mol);Fe:y(mol)`

`27x+56y=11(1)`

`2Al+6HCl->2AlCl_3+3H_2`

`Fe+2HCl->FeCl_2+H_2`

Theo pt:

`n_{H_2}=1,5x+y={8,96}/{22,4}=0,4(2)`

`(1)(2)->x=0,2;y=0,1`

`m_{Al}=0,2.27=5,4(g);m_{Fe}=0,1.56=5,6(g)`

`b)`

Theo pt:

`n_{HCl}=2n_{H_2}=0,8(mol)`

`C\%_{HCl}={0,8.36,5}/{500}.100\%=5,84\%`