Em tham khảo nha :
\(\begin{array}{l}
a)\\
{n_X} = \dfrac{V}{{22,4}} = \dfrac{{13,44}}{{22,4}} = 0,6mol\\
{M_X} = \dfrac{m}{n} = \dfrac{{17,6}}{{0,6}} = 29,33dvC\\
{d_{X/{H_2}}} = \dfrac{{{M_X}}}{{{M_{{H_2}}}}} = \dfrac{{29,33}}{2} = 14,665\\
b)\\
{n_X} = \dfrac{m}{M} = \dfrac{{3,3}}{{29,33}} = 0,1125mol\\
{n_{{H_2}}} = {n_X} = 0,1125mol\\
{V_{{H_2}}} = n \times 22,4 = 0,1125 \times 22,4 = 2,52l\\
c)\\
2CO + {O_2} \to 2C{O_2}\\
hh:{N_2}(a\,mol),{O_2}(b\,mol)\\
a + b = 0,6(1)\\
28a + 32b = 17,6(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,4;b = 0,2\\
{n_{C{O_2}}} = 2{n_{{O_2}}} = 0,4mol\\
{m_{C{O_2}}} = n \times M = 0,4 \times 44 = 17,6g\\
{m_{{N_2}}} = n \times M = 0,4 \times 28 = 11,2g\\
\% C{O_2} = \dfrac{{17,6}}{{17,6 + 11,2}} \times 100\% = 61,1\% \\
\% {N_2} = 100 - 61,1 = 38,9\% \\
\% {V_{C{O_2}}} = \dfrac{{0,4 \times 22,4}}{{(0,4 + 0,4) \times 22,4}} \times 100\% = 50\% \\
\% {V_{{N_2}}} = 100 - 50 = 50\%
\end{array}\)
