Em tham khảo nha :
\(\begin{array}{l}
a)\\
C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\\
{V_{C{H_4}}} = 2l\\
{n_{{O_2}}} = 2{n_{C{H_4}}}\\
\Rightarrow {V_{{O_2}}} = 2{V_{C{H_4}}} = 4l\\
b)\\
{n_{C{O_2}}} = {n_{C{H_4}}} = 0,15mol\\
{V_{C{O_2}}} = 0,15 \times 22,4 = 3,36l\\
c)\\
{d_{C{H_4}/kk}} = \dfrac{{16}}{{29}} = 0,552\\
\text{Khí metan nhẹ hơn không khí 0,552 lần}
\end{array}\)