Em tham khảo nha :
\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = \dfrac{{5,6}}{{56}} = 0,1mol\\
{n_{HCl}} = 0,3 \times 1 = 0,3mol\\
\dfrac{{0,1}}{1} < \dfrac{{0,3}}{2} \Rightarrow HCl\text{ dư}\\
{n_{{H_2}}} = {n_{Fe}} = 0,1mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
{n_{HC{l_d}}} = 0,3 - 0,2 = 0,1mol\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1mol\\
{C_{{M_{FeC{l_2}}}}} = \dfrac{{0,1}}{{0,3}} = \dfrac{1}{3}M\\
{C_{{M_{HC{l_d}}}}} = \dfrac{{0,1}}{{0,3}} = \dfrac{1}{3}M
\end{array}\)