\(n_{Fe}=0,1\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0,1\) \(0,2\) \(0,1\)
\(m_{ddHCl}=\dfrac{0,2.36,5}{50\%}=36,5\left(g\right)\)
\(m_{H_2}=0,1.2=0,2\left(g\right)\)
\(m_{dd}=5,6+36,5-0,2=41,9\left(g\right)\)
\(C\%\left(FeCl_2\right)=\dfrac{0,1.127}{41,9}=30,3\left(\%\right)\)