Em tham khảo nha:

\(\begin{array}{l}
c)\\
{n_P} = \dfrac{{0,62}}{{31}} = 0,02\,mol\\
4P + 5{O_2} \xrightarrow{t^0} 2{P_2}{O_5}\\
{n_{{O_2}}} = 0,02 \times \dfrac{5}{4} = 0,025\,mol\\
{V_{{O_2}}} = 0,025 \times 22,4 = 0,56l\\
{m_{{O_2}}} = 0,025 \times 32 = 0,8g\\
d)\\
{n_{CO}} = \dfrac{{14}}{{28}} = 0,5\,mol\\
2CO + {O_2} \xrightarrow{t^0} 2C{O_2}\\
{n_{{O_2}}} = 0,5 \times \dfrac{1}{2} = 0,25\,mol\\
{V_{{O_2}}} = 0,25 \times 22,4 = 5,6l\\
{m_{{O_2}}} = 0,25 \times 32 = 8g\\
e)\\
{n_{Al}} = \dfrac{{6,75}}{{27}} = 0,25\,mol\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
{n_{{O_2}}} = 0,25 \times \dfrac{3}{4} = 0,1875\,mol\\
{V_{{O_2}}} = 0,1875 \times 22,4 = 4,2l\\
{m_{{O_2}}} = 0,1875 \times 32 = 6g\\
f)\\
{n_{{C_4}{H_8}{O_2}}} = \dfrac{{8,8}}{{88}} = 0,1\,mol\\
{C_4}{H_8}{O_2} + 5{O_2} \xrightarrow{t^0} 4C{O_2} + 4{H_2}O\\
{n_{{O_2}}} = 0,1 \times 5 = 0,5\,mol\\
{V_{{O_2}}} = 0,5 \times 22,4 = 11,2l\\
{m_{{O_2}}} = 0,5 \times 32 = 16g
\end{array}\)

Đáp án:

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