Em tham khảo nha:

\(\begin{array}{l}
4)\\
a)\\
{n_{{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
2Cu + {O_2} \xrightarrow{t^0} 2CuO\\
{n_{CuO}} = 2{n_{{O_2}}} = 0,4\,mol\\
{m_{CuO}} = 0,4 \times 80 = 32g\\
b)\\
{n_{Cu}} = 2{n_{{O_2}}} = 0,4\,mol\\
{m_{Cu}} = 0,4 \times 64 = 25,6g\\
c)\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = 0,2 \times \dfrac{2}{3} = \dfrac{2}{{15}}\,mol\\
{m_{KCl{O_3}}} = \dfrac{2}{{15}} \times 22,4 = 2,9867g\\
5)\\
a)\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = \dfrac{{49}}{{122,5}} = 0,4\,mol\\
{n_{{O_2}}} = 0,4 \times \frac{3}{2} = 0,6\,mol\\
{V_{{O_2}}} = 0,6 \times 22,4 = 13,44l\\
b)\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
{n_{C{H_4}}} = \dfrac{{11,2}}{{22,4}} = 0,5\,mol\\
\text{ Lập tỉ lệ }:\dfrac{{{n_{C{H_4}}}}}{1} > \dfrac{{{n_{{O_2}}}}}{2}(0,5 > 0,3)\\
 \Rightarrow \text{ $CH_4$ dư }
\end{array}\)