Em tham khảo nha :
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{HCl}} = 0,25 \times 1 = 0,25mol\\
{n_{Fe}} = \dfrac{{{n_{HCl}}}}{2} = 0,125mol\\
{m_{Fe}} = 0,125 \times 56 = 7g\\
b)\\
{n_{{H_2}}} = \dfrac{{{n_{HCl}}}}{2} = 0,125mol\\
{V_{{H_2}}} = 0,125 \times 22,4 = 2,8l
\end{array}\)
