Em tham khảo nha :
\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
b)\\
{n_{Fe}} = \dfrac{{28}}{{56}} = 0,5mol\\
{n_{{H_2}}} = {n_{Fe}} = 0,5mol\\
{V_{{H_2}}} = 0,5 \times 22,4 = 11,2l\\
c)\\
{n_{CuO}} = \dfrac{{12}}{{80}} = 0,15mol\\
CuO + {H_2} \to Cu + {H_2}O\\
\dfrac{{0,15}}{1} < \dfrac{{0,5}}{1} \Rightarrow {H_2}\text{ dư}\\
{n_{{H_2}d}} = {n_{{H_2}}} - {n_{CuO}} = 0,5 - 0,15 = 0,35mol\\
{m_{{H_2}d}} = 0,35 \times 2 = 0,7g
\end{array}\)