Em tham khảo nha:
\(\begin{array}{l}
a)\\
4P + 5{O_2} \xrightarrow{t^0} 2{P_2}{O_5}\\
{n_P} = \dfrac{m}{M} = \dfrac{{46,5}}{{31}} = 1,5\,mol\\
{n_{{O_2}}} = 1,5 \times \dfrac{5}{4} = 1,875\,mol\\
{V_{{O_2}}} = n \times 22,4 = 1,875 \times 22,4 = 42l\\
b)\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
{n_{Al}} = \dfrac{m}{M} = \dfrac{{67,5}}{{27}} = 2,5\,mol\\
{n_{{O_2}}} = 2,5 \times \dfrac{3}{4} = 1,875\,mol\\
{V_{{O_2}}} = n \times 22,4 = 1,875 \times 22,4 = 42l\\
c)\\
2{H_2} + {O_2} \xrightarrow{t^0} 2{H_2}O\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{33,6}}{{22,4}} = 1,5\,mol\\
{n_{{O_2}}} = 1,5 \times \dfrac{1}{2} = 0,75\,mol\\
{V_{{O_2}}} = n \times 22,4 = 0,75 \times 22,4 = 16,8l
\end{array}\)
