$\begin{align}
& {{n}_{{{H}_{2}}}}=5,6:22,4=0,25(mol) \\
& {{n}_{C{{l}_{2}}}}=3,36:22,4=0,15(mol) \\
\end{align}$
a)
$\begin{align}
& {{H}_{2}}+C{{l}_{2}}\to 2HCl \\
& x\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x \\
\end{align}$
Bảo toàn khối lượng
${{m}_{A}}={{m}_{HCl}}+{{m}_{{{H}_{2}}O}}=36,5\times 2x+192,7=73x+192,7$
$HCl+AgN{{O}_{3}}\to AgCl\downarrow +HN{{O}_{3}}$
${{n}_{AgCl}}=7,175:143,5=0,05(mol)={{n}_{HCl}}$
Trong 73x + 192,7 g dung dịch A có 2x mol HCl
Trong 50g A có 0,05mol HCl
Ta có:
$0,05\times (73x+192,7)=2x\times 50$
x = 0,1mol
Hiệu suất tính theo $C{{l}_{2}}$
H = $\frac{0,1}{0,15}\times 100%=66,67%$
b) Trong 100g dung dịch A có: ${{n}_{HCl}}=0,05\times 2=0,1(mol)$
${{m}_{muoi}}={{m}_{KL}}+{{m}_{C{{l}^{-}}}}=1,3+0,1\times 35,5=4,85g>3,9625g$
$\begin{align}
& M+2HCl\to MC{{l}_{2}}+{{H}_{2}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a \\
\end{align}$
Bảo toàn khối lượng:
$\begin{align}
& {{m}_{M}}+{{m}_{HCl}}={{m}_{r}}+{{m}_{{{H}_{2}}}} \\
& 1,3+36,5\times 2a=3,9625+2a \\
& a=0,0375mol \\
\end{align}$
$\left\{ \begin{align}
& {{n}_{Fe}}+{{n}_{Mg}}=0,0375 \\
& 56\times {{n}_{Fe}}+24\times {{n}_{Mg}}=1,3 \\
\end{align} \right.$
$\begin{align}
& {{n}_{Fe}}=0,0125={{n}_{FeC{{l}_{2}}}} \\
& {{n}_{Mg}}=0,025={{n}_{MgC{{l}_{2}}}} \\
\end{align}$
% ${{m}_{FeC{{l}_{2}}}}=\frac{0,0125\times 127}{3,9625}\times 100$% = 40,06%
% ${{m}_{MgC{{l}_{2}}}}=$100% - 40,06% = 59,94%