mình cần giải thích rõ ra ạ

Đáp án `+` Giải thích các bước giải:

`a)`

`Fe + 2HCl -> FeCl_2 + H_2`

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

`b)`

Đặt `n_{Fe} = a(mol); n_{Al} = b(mol)`

`-> 56a + 27b = 11(1)`

`n_{H_2} = (8,96)/(22,4) = 0,4(mol)`

Theo PT: `n_{H_2} = n_{Fe} + 3/2 n_{Al}`

`-> a + 1,5b = 0,4(2)`

`(1),(2) -> a = 0,1; b = 0,2`

`\%m_{Al} = (0,1.27)/(11) .100\% = 24,55\%`

`\%m_{Fe} = 100\% - 24,55\% = 75,45\%`

`c)`

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

`-> m_{ddHCl} = (0,8.36,5)/(3,65\%) = 800(g)`

`-> m_{dd(sau)} = 800 + 11 - 0,4.2 = 810,2(g)`

Theo PT:

`n_{FeCl_2} = n_{Fe} = 0,1(mol)`

`n_{AlCl_3} = n_{Al} = 0,2(mol)`

`C\%_{FeCl_2} = (0,1.127)/(810,2) .100\% = 1,57\%`

`C\%_{AlCl_3} = (0,2.133,5)/(810,2) .100\% = 3,3\%`

`a)`

`Fe+2HCl-.FeCl_2+H_2`

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

`b)`

Đặt `n_{Fe}=x(mol);n_{Al}=y(mol)`

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

Theo PT: `x+1,5y=n_{H_2}=0,4(2)`

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

`\%m_{Fe}={0,1.56}/{11}.100\%=50,91\%`

`\%m_{Al}=100-50,91=49,09\%`

`c)`

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

`->m_{dd\ HCl}={0,8.36,5}/{3,65\%}=800(g)`

`->m_{dd\ spu}=11+800-0,4.2=810,2(g)`

Theo PT: `n_{FeCl_2}=0,1(mol);n_{AlCl_3}=0,2(mol)`

`C\%_{FeCl_2}={0,1.152}/{810,2}.100\%\approx 1,88\%`

`C\%_{AlCl_3}={0,2.133,5}/{810,2}.100\%\approx 3,295\%`