a.
`A+2HCl→ACl_2+H_2`
`2B+6HCl→2BCl_3+3H_2`
b. `n_{H_2}=\frac{3,808}{22,4}=0,17` $(mol)$
Ta thấy: $n_{HCl}=2.n_{H_2}=2.0,17=0,34$ $(mol)$
Theo ĐLBTKL ta có:
$m_{hh}+m_{HCl}=m_{muối}+m_{H_2}$
$⇒m_{muối}=m_{hh}+m_{HCl}-m_{H_2}$
$⇒m_{muối}=4+0,34.36,5-0,17.2=16,07$ $(g)$