Đáp án:

`m_{NaCl}=58,5g`

Giải thích các bước giải:

`a) 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2`

`Fe + H_2SO_4 \to FeSO_4 + H_2`

Theo pt: `n_{H_2}=1,5n_{Al}+n_{Fe}={8,96}/{22,4}=0,4 mol`

=> `{(1.5n_{Al}+n_{Fe}=0.4),(27n_{Al}+56n_{Fe}=11):}`

`=> n_{Al}=0,2, n_{Fe}=0,1 mol`

`=> m_{Al}=0,2.27=5,4g`

`m_{Fe}=56.0,1=5,6g`

$b) 2NaCl + 2H_2O \xrightarrow{đpddcmn} 2NaOH + Cl_2 + H_2$

Theo pt: `n_{NaCl}={2.n_{H_2}}/{H%}={2.0,4}/{0,8}=1 mol`

`=> m_{NaCl}=58,5g`