$n_{HCl}=\dfrac{14,6}{36,5}=0,4mol$
$PTHH :$
$MgO+2HCl\to MgCl_2+H_2↑(1)$
$ZnO+2HCl\to ZnCl_2+H_2↑(2)$
$Gọi\ n_{MgO}=a;n_{ZnO}=b(a,b>0)$
$Ta\ có : \\m_{hh}=40a+81b=12,1g \\n_{HCl}=2a+2b=0,4mol$
Ta có hpt :
$\left\{\begin{matrix} 40a+81b=12,1 & \\ 2a+2b=0,4 & \end{matrix}\right. ⇔\left\{\begin{matrix} a=0,1 & \\ b=0,1 & \end{matrix}\right.$
$\text{Theo pt (1) và (2) :}$
$n_{MgCl_2}=n_{Mg}=0,1mol$
$n_{ZnCl_2}=n_{Zn}=0,1mol$
$⇒a=m_{muối}=m_{MgCl_2}+m_{ZnCl_2}=0,1.95+0,1.136=23,1g$