Đáp án:
\(\begin{array}{l}
a)\\
{V_{{H_2}}} = 4,48l\\
b)\\
{m_{Cu}} = 12,8g\\
c)\\
{m_{Cu}} = 9,6g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
CuO + {H_2} \to Cu + {H_2}O\\
a)\\
{n_{CuO}} = \dfrac{m}{M} = \dfrac{{16}}{{80}} = 0,2mol\\
{n_{{H_2}}} = {n_{Cu}} = 0,2mol\\
{V_{{H_2}}} = n \times 22,4 = 0,2 \times 22,4 = 4,48l\\
b)\\
{n_{Cu}} = {n_{CuO}} = 0,2mol\\
{m_{Cu}} = n \times M = 0,2 \times 64 = 12,8g\\
c)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
\dfrac{{0,15}}{1} < \dfrac{{0,2}}{1} \Rightarrow CuO\text{ dư}\\
{n_{Cu}} = {n_{{H_2}}} = 0,15mol\\
{m_{Cu}} = n \times M = 0,15 \times 64 = 9,6g
\end{array}\)