Em tham khảo nha :
\(\begin{array}{l}
3CuO + 2N{H_3} \to 3Cu + {N_2} + 3{H_2}O\\
{n_{N{H_3}}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
{n_{CuO}} = \dfrac{{16}}{{80}} = 0,2mol\\
\dfrac{{0,2}}{3} > \dfrac{{0,05}}{2} \Rightarrow CuO\text{ dư}\\
{n_{Cu{O_d}}} = 0,2 - \frac{3}{2} \times 0,05 = 0,125mol\\
{m_{Cu{O_d}}} = 0,125 \times 80 = 10g\\
{n_{Cu}} = \dfrac{3}{2}{n_{N{H_3}}} = 0,075mol\\
{m_{Cu}} = 0,075 \times 64 = 4,8g\\
{m_X} = 10 + 4,8 = 14,8g\\
b)\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
{n_{HCl}} = 2{n_{Cu{O_d}}} = 0,25mol\\
{V_{HCl}} = \dfrac{{0,25}}{{0,5}} = 0,5l
\end{array}\)