1.
$n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)$
Đặt $x$, $y$ là số mol $Fe$, $M$
Bảo toàn e: $2n_{Fe}+2n_M=2n_{H_2}$
$\to x+y=0,4$ $(1)$
$n_{Cl_2}=\dfrac{10,64}{22,4}=0,475(mol)$
Bảo toàn e: $3n_{Fe}+2n_M=2n_{Cl_2}$
$\to 3x+2y=0,475.2=0,95$ $(2)$
$(1)(2)\to x=0,15; y=0,25$
Muối clorua ở TN1 gồm $FeCl_2$ ($0,15$ mol), $MCl_2$ ($0,25$ mol)
$\to 0,15.127+0,25(M_M+71)=53,05$
$\to M_M=65(Zn)$
$\to \%m_{Zn}=\dfrac{0,25.65.100}{0,15.56+0,25.65}=65,92\%$
2.
$n_{AgBr\downarrow}=\dfrac{28,2}{188}=0,15(mol)$
$\to n_{NaBr}=n_{AgBr}=0,15(mol)$ (bảo toàn $Br$)
$n_{NaF}:n_{NaBr}=1:1\to n_{NaF}=0,15(mol)$
$\to m=0,15.42+0,15.103=21,75g$