Em tham khảo nha:

\(\begin{array}{l}
a)\\
\text{ Giả sử số mol HCl là 1 mol } \\
{n_{{O_2}}} = 4{n_{HCl}} = 4\,mol\\
\% {V_{HCl}} = \dfrac{1}{{1 + 4}} \times 100\%  = 20\% \\
\% {V_{{O_2}}} = 100 - 20 = 80\% \\
\% {m_{HCl}} = \dfrac{{1 \times 36,5}}{{1 \times 36,5 + 4 \times 32}} \times 100\%  = 22,19\% \\
\% {m_{{O_2}}} = 100 - 22,19 = 77,81\% \\
b)\\
{M_A} = \dfrac{{1 \times 36,5 + 4 \times 32}}{{1 + 4}} = 32,9g/mol\\
{d_{A/kk}} = \dfrac{{32,9}}{{29}} = 1,13\\
c)\\
{M_{hh}} = 16,75 \times 2 = 33,5g/mol\\
hh:{O_2}(a\,mol),HCl(b\,mol)\\
\text{ Giả sử số mol hỗn hợp là 1 mol } \\
\left\{ \begin{array}{l}
a + b = 1\\
32a + 36,5b = 33,5
\end{array} \right.\\
 \Rightarrow a = \dfrac{2}{3};b = \dfrac{1}{3}\\
{n_{HCl}}:{n_{{O_2}}} = \dfrac{1}{3}:\dfrac{2}{3} = 1:2
\end{array}\)