Em tham khảo nha :
\(\begin{array}{l}
1)\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
{n_{C{l_2}}} = \dfrac{{44,8}}{{22,4}} = 0,2mol\\
{n_{Fe}} = \dfrac{2}{3}{n_{C{l_2}}} = \dfrac{2}{{15}}mol\\
{m_{Fe}} = \dfrac{2}{{15}} \times 56 = 7,47g\\
{n_{FeC{l_3}}} = {n_{Fe}} = \dfrac{2}{{15}}mol\\
{m_{FeC{l_3}}} = \dfrac{2}{{15}} \times 162,5 = 21,67g\\
2)\\
{H_2} + C{l_2} \to 2HCl\\
{n_{{H_2}}} = \dfrac{{24,4}}{{22,4}} = 1,089mol\\
{n_{C{l_2}}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
\dfrac{{0,05}}{1} < \dfrac{{1,089}}{1} \Rightarrow {H_2}\\
{n_{HCl}} = 2{n_{C{l_2}}} = 0,1mol\\
{V_{HCl}} = 0,1 \times 22,4 = 2,24l
\end{array}\)