Em tham khảo nha :

\(\begin{array}{l}
C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\\
{n_{C{H_4}}} = \dfrac{V}{{22,4}} = \dfrac{{11,2}}{{22,4}} = 0,5mol\\
{n_{{O_2}}} = 2{n_{C{H_4}}} = 1mol\\
{V_{{O_2}}} = n \times 22,4 = 1 \times 22,4 = 22,4l\\
{V_{kk}} = \dfrac{{22,4 \times 100}}{{20}} = 112l\\
{n_{C{O_2}}} = {n_{C{H_4}}} = 0,5mol\\
{V_{C{O_2}}} = n \times 22,4 = 0,5 \times 22,4 = 11,2l\\
{n_{{H_2}O}} = 2{n_{C{H_4}}} = 1mol\\
{m_{{H_2}O}} = n \times M = 1 \times 18 = 18g\\
2)\\
2Cu + {O_2} \to 2CuO\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{Cu}} = \dfrac{m}{M} = \dfrac{{60}}{{64}} = 0,9375mol\\
\dfrac{{0,9375}}{2} > \dfrac{{0,2}}{1} \Rightarrow Cu\text{ dư}\\
2{n_{CuO}} = {n_{{O_2}}} = 0,4mol\\
{m_{CuO}} = n \times M = 0,4 \times 80 = 32g\\
{n_{C{u_{pu}}}} = 2{n_{{O_2}}} = 0,4mol\\
{m_{C{u_{pu}}}} = n \times M = 0,4 \times 64 = 25,6g\\
{m_{cr}} = 60 - 25,6 + 32 = 66,4g\\
\% CuO = \dfrac{{32}}{{66,4}} \times 100\%  = 48,19\% 
\end{array}\)