Đáp án:

 `86,1(g)`

Giải thích các bước giải:

$\begin{cases}A:Fe\\B:Zn\end{cases}+HCl \begin{cases}FeCl_2\\ZnCl_2\end{cases}+AgNO_3 \to AgCl$

Đặt :

$n_{Fe}=n_{FeCl_2}=x(mol)$

$n_{Zn}=n_{ZnCl_2}=y(mol)$

$\Rightarrow \begin{cases}56x + 65y=18,6\\127x+136y=39,9\end{cases} \Leftrightarrow \begin{cases}x=0,1(mol)\\y=0,2(mol)\end{cases}$

$FeCl_2 + 2AgNO_3 \to Fe(NO_3)_2 + 2AgCl\\ZnCl_2 + 2AgNO_3 \to Zn(NO_3)_2 + 2AgCl$

`->\sum n_{AgCl}=0,2xx2+0,1xx2=0,6(mol)`

`=>m_{AgCl}=0,6xx143,5=86,1(g)`

`text{# Như}`

`#Ly`

PTHH:

$Fe$ $+$ $2HCl$ `->` $FeCl_2$ $+$ $H_2$

$Zn$ $+$ $2HCl$ `->` $ZnCl_2$ $+$ $H_2$

$FeCl_2$ $+$ $2AgNO_3$ `->` $Fe(NO_3)_2$ $+$ $2AgCl$

$ZnCl_2$ $+$ $2AgNO_3$ `->` $Zn(NO_3)_2$ $+$ $2AgCl$

Ta có:

`m_(KL` `=` `m_(KL(\text{Muối})` `=` `18,6` $(g)$

`m_(Cl_(\text{Muối})` `=` `39,9-18,6=21,3` $(g)$

`n_(Cl_` `=` `(21,3)/(35,5)=0,6` $(mol)$

Bảo toàn $Cl:$ `n_(AgCl` `=` `n_(Cl_` `=`  `0,6` $(mol)$

`m_(AgCl` `=` `0,6.143,5=86,1` $(g)$