Em tham khảo nha :
\(\begin{array}{l}
4)\\
a)\\
4P + 5{O_2} \xrightarrow{t^0} 2{P_2}{O_5}\\
b)\\
{n_{{P_2}{O_5}}} = \dfrac{{28,4}}{{142}} = 0,2mol\\
{n_P} = 2{n_{{P_2}{O_5}}} = 0,4mol\\
{m_P} = n \times M = 0,4 \times 31 = 12,4g\\
c)\\
{n_{{O_2}}} = \dfrac{5}{2}{n_{{P_2}{O_5}}} = 0,5mol\\
{V_{{O_2}}} = n \times 22,4 = 0,5 \times 22,4 = 11,2l\\
5)\\
a)\\
2Cu + {O_2} \xrightarrow{t^0} 2CuO\\
b)\\
{n_{Cu}} = \dfrac{m}{M} = \dfrac{{16,8}}{{64}} = 0,2625mol\\
{n_{{O_2}}} = \dfrac{{{n_{Cu}}}}{2} = 0,13125mol\\
{V_{{O_2}}} = 0,13125 \times 22,4 = 2,94l\\
{n_{CuO}} = {n_{Cu}} = 0,2625mol\\
{m_{CuO}} = 0,2625 \times 80 = 21g\\
c)\\
2KMn{O_4} \xrightarrow{t^0} {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
{n_{KMn{O_4}}} = 2{n_{{O_2}}} = 0,2625mol\\
{m_{KMn{O_4}}} = 0,2625 \times 158 = 41,475g\\
6)\\
a)\\
3Fe + 2{O_2} \xrightarrow{t^0} F{e_3}{O_4}\\
{n_{F{e_3}{O_4}}} = \dfrac{{4,64}}{{232}} = 0,02mol\\
{n_{Fe}} = 3{n_{F{e_3}{O_4}}} = 0,06mol\\
{m_{Fe}} = 0,06 \times 56 = 3,36g\\
{n_{{O_2}}} = 2{n_{F{e_3}{O_4}}} = 0,04mol\\
{m_{{O_2}}} = 0,04 \times 32 = 1,28g\\
b)\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = \dfrac{2}{3}{n_{{O_2}}} = \dfrac{2}{{75}}mol\\
{m_{KCl{O_3}}} = \dfrac{2}{{75}} \times 122,5 = 3,267g
\end{array}\)
