Em tham khảo nha :
\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
b)\\
{n_{Zn}} = \dfrac{m}{M} = \dfrac{{13}}{{65}} = 0,2mol\\
{n_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
c)\\
CuO + {H_2} \to Cu + {H_2}O\\
{n_{CuO}} = \dfrac{{12}}{{80}} = 0,15mol\\
\dfrac{{0,15}}{1} < \dfrac{{0,2}}{1} \Rightarrow {H_2}\text{ dư}\\
{n_{{H_2}d}} = {n_{{H_2}}} - {n_{CuO}} = 0,2 - 0,15 = 0,05mol\\
{V_{{H_2}d}} = 0,05 \times 22,4 = 1,12l
\end{array}\)