Em tham khảo nha :
\(\begin{array}{l}
{n_{N{O_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{{N_2}}} = \dfrac{{5,6}}{{28}} = 0,2mol\\
{M_B} = \dfrac{{0,1 \times 46 + 0,2 \times 28}}{{0,3}} = 34dvC\\
{n_B} = \dfrac{1}{{22,4}} = \dfrac{5}{{112}}mol\\
{m_B} = \dfrac{5}{{112}} \times 34 = 1,52g\\
\% N{O_2} = \dfrac{{0,1 \times 46}}{{0,1 \times 46 + 5,6}} \times 100\% = 45,1\% \\
\% {N_2} = 100 - 45,1 = 54,9\%
\end{array}\)