$a)\\ Fe+2HCl\xrightarrow\ FeCl_2+H_2\\ Mg+2HCl\xrightarrow\ MgCl_2+H_2\\ Zn+2HCl\xrightarrow\ ZnCl_2+H_2\\ b)\\ m_{Cl^{-}}=m+7,1-m=7,1(g)\\ n_{HCl(pu)}=n_{Cl^{-}}=\dfrac{7,1}{71}=0,1(mol)\\ n_{HCl}=2n_{H_2}\ (1)\\  Fe+H_2SO_4\xrightarrow\ FeSO_4+H_2\\ Mg+H_2SO_4\xrightarrow\ MgSO_4+H_2\\ Zn+H_2SO_4\xrightarrow\ ZnSO_4+H_2\\ n_{H_2SO_4}=n_{H_2}\ (2)\\ (1)(2)\xrightarrow\ n_{SO_4^{2-}}=n_{H_2SO_4}=\dfrac{1}{2}n_{HCl}=0,05(mol)\\ m_{muoi}=m+0,05.96=m+4,8(g)$

`a)`

`Fe+2HCl->FeCl_2+H_2`

`Mg+2HCl->MgCl_2+H_2`

`Zn+2HCl->ZnCl_2+H_2`

`b)`

BTKL: `m_{HCl}-m_{H_2}=m_{m u o i}-m_{KL}`

`->36,5n_{HCl}-2n_{H_2}=m+7,1-m=7,1`

Mà `n_{HCl}=2n_{H_2}`

`->n_{HCl}=0,2(mol)`

`Fe+H_2SO_4->FeSO_4+H_2`

`Mg+H_2SO_4->MgSO_4+H_2`

`Zn+H_2SO_4->ZnSO_4+H_2`

Theo pt:

`n_{H_2}=n_{H_2SO_4}=1/2n_{HCl}=0,1(mol)`

BTKL: `m+0,1.98=m_{m u o i}+0,1.2`

`->m_{m u o i}=m+9,6(g)`