$n_{H_2}=6,72/22,4=0,3mol$
$PTHH :$
$Fe+2HCl\to FeCl_2+H_2↑$
$Zn+2HCl\to ZnCl_2+H_2↑$
Gọi $n_{Fe}=a;n_{Zn}=b$
Ta có :
$m_{hh}=56a+65b=17,7g$
$n_{H_2}=a+b=0,3mol$
Ta có hpt :
$\left\{\begin{matrix} 56a+65b=17,7 & \\ a+b=0,3 & \end{matrix}\right.$ $⇔\left\{\begin{matrix} a=0,2 & \\ b=0,1 & \end{matrix}\right.$
$⇒m_{Fe}=0,2.56=11,2g$
$m_{Zn}=17,7-11,2=6,5g$