\(\begin{array}{l}
a)\\
KL:\,R\\
R+2HCl\to RCl_2+H_2\\
n_R=n_{H_2}=\frac{4,48}{22,4}=0,2(mol)\\
M_R=\frac{11,2}{0,2}=56(g/mol)\\
\to R:\,sat\,(Fe)\\
b)\\
n_{HCl}=2n_{H_2}=0,4(mol)\\
m_{dd\,HCl}=\frac{0,4.36,5}{14,6\%}=100(g)\\
c)\\
n_{FeCl_2}=n_{Fe}=0,2(mol)\\
C\%_{FeCl_2}=\frac{0,2.127}{11,2+100-0,2.2}.100\%\approx 22,92\%
\end{array}\)