Em tham khảo nha :

\(\begin{array}{l}
a)\\
{n_X} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
hh:{H_2}(a\,mol),{O_2}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,05\\
2a + 32b = 0,55
\end{array} \right.\\
 \Rightarrow a = 0,035;b = 0,015\\
\% {H_2} = \dfrac{{0,035 \times 22,4}}{{1,12}} \times 100\%  = 70\% \\
\% {O_2} = 100 - 70 = 30\% \\
b)\\
{M_X} = \dfrac{{0,55}}{{0,05}} = 11dvC\\
{d_{X/kk}} = \dfrac{{11}}{{29}} = 0,379\\
\text{Hỗn hợp khí A nhẹ hơn không khí }\\
c)\\
2{H_2} + {O_2} \to 2{H_2}O\\
{V_{{H_2}}} = \dfrac{{13,44 \times 70}}{{100}} = 9,408l\\
{n_{{H_2}}} = \dfrac{{0,42}}{{22,4}} = 0,42mol\\
{n_{{O_2}}} = \dfrac{{13,44}}{{22,4}} - 0,42 = 0,18mol\\
\dfrac{{0,42}}{2} > \dfrac{{0,18}}{1} \Rightarrow {H_2}\text{ dư}\\
{n_{{H_2}O}} = 2{n_{{O_2}}} = 0,36mol\\
{m_{{H_2}O}} = 0,36 \times 18 = 6,48g\\
{V_{{H_2}O}} = {m_{{H_2}O}} \times 1 = 6,48ml
\end{array}\)