Em tham khảo nha:
\(\begin{array}{l}
\text{ Trước phản ứng }\\
{M_{hh}} = \dfrac{{a \times 32 + 3a \times 64}}{{a + 3a}} = 56g/mol\\
\Rightarrow {d_{hh/{O_2}}} = \dfrac{{56}}{{32}} = 1,75\\
\text{ Sau phản ứng }\\
2S{O_2} + {O_2} \xrightarrow{t^0,V_2O_5 } 2S{O_3}\\
\dfrac{{{n_{S{O_2}}}}}{2} > {n_{{O_2}}} \Rightarrow \text{ $SO_2$ dư } \\
{n_{S{O_2}}} \text{ dư }= 3a - 2 \times a = a\,mol\\
{n_{S{O_3}}} = 2{n_{{O_2}}} = 2a\,mol\\
{M_{hh}} = \dfrac{{a \times 32 + 2a \times 80}}{{a + 2a}} = 64g/mol\\
{d_{hh/{O_2}}} = \dfrac{{64}}{{32}} = 2
\end{array}\)