Em tham khảo nha:

\(\begin{array}{l}
a,b)\\
S + {O_2} \xrightarrow{t^0} S{O_2}(oxit\,axit: \text{ Lưu huỳnh dioxit }\\
4P + 5{O_2} \xrightarrow{t^0} 2{P_2}{O_5}(oxit\,axit:diphotpho\,pentaoxit)\\
2Mg + {O_2} \xrightarrow{t^0} 2MgO(\,oxit\,bazo:\,Magie\,oxit)\\
{C_3}{H_8} + 5{O_2} \xrightarrow{t^0} 3C{O_2}(oxit\,axit:\,cacbon\,dioxit) + 4{H_2}O\\
2Cu + {O_2} \xrightarrow{t^0} 2CuO(o\,xit\,bazo: \text{ đồng (II) oxit }\\
4Na + {O_2} \xrightarrow{t^0} 2N{a_2}O(\,o\,xit\,bazo:Natri\,oxit)\\
c)\\
S + {O_2} \xrightarrow{t^0} S{O_2}\\
{n_S} = \dfrac{{6,4}}{{32}} = 0,2\,mol\\
{n_{{O_2}}} = {n_S} = 0,2\,mol\\
{V_{{O_2}}} = 0,2 \times 22,4 = 4,48l\\
d)\\
2KMn{O_4} \xrightarrow{t^0} {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
{n_{KMn{O_4}}} = \dfrac{{15,8}}{{158}} = 0,1\,mol\\
{n_{{O_2}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
2Mg + {O_2} \xrightarrow{t^0} 2MgO\\
{n_{Mg}} = \dfrac{{12}}{{24}} = 0,5\,mol\\
{n_{Mg}}\text{ dư } = 0,5 - \dfrac{{0,05}}{2} = 0,475\,mol\\
{n_{MgO}} = \dfrac{{0,05}}{2} = 0,025\,mol\\
\% {m_{Mg}} \text{ dư }= \dfrac{{0,475 \times 24}}{{0,475 \times 24 + 0,025 \times 40}} \times 100\%  = 91,94\% \\
\% {m_{MgO}} = 100 - 91,94 = 8,06\% 
\end{array}\)