Đáp án:

\(\begin{array}{l}
a)\\
{m_{HC{l_d}}} = 0,1825g\\
b)\\
{V_{{H_2}}} = 2,24l
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = \dfrac{m}{M} = \dfrac{{5,6}}{{56}} = 0,1mol\\
{n_{HCl}} = \dfrac{m}{M} = \dfrac{{7,5}}{{36,5}} = 0,205mol\\
\dfrac{{0,1}}{1} < \dfrac{{0,205}}{2} \Rightarrow HCl\text{ dư}\\
{n_{HC{l_d}}} = {n_{HCl}} - 2{n_{Fe}} = 0,005mol\\
{m_{HC{l_d}}} = n \times M = 0,005 \times 36,5 = 0,1825g\\
b)\\
{n_{{H_2}}} = {n_{Fe}} = 0,1mol\\
{V_{{H_2}}} = n \times 22,4 = 0,1 \times 22,4 = 2,24l
\end{array}\)