\(a.n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,2 0,1 0,1
\(m_{Fe}=0,1.56=5,6g\\ m_{FeO}=13,6-5,6=8g\)
\(b.n_{FeO}=\dfrac{8}{72}=\dfrac{1}{9}mol\)
\(FeO+2HCl\rightarrow FeCl_2+H_2O\)
\(\dfrac{1}{9}\) \(\dfrac{2}{9}\) \(\dfrac{1}{9}\)
\(C_{M_{HCl}}=\dfrac{0,2+\dfrac{2}{9}}{0,2}=\dfrac{19}{9}M\)
\(c.m_{FeCl_2}=\left(0,1+\dfrac{1}{9}\right)127=26,81g\)