$\rm M_{X} = 2 \times 28 = 56g/mol$

Quy tắc đường chéo:

 $\rm {M_{SO_2}=64}\hspace{4cm}{24}\\\hspace{1,4cm}\huge{\searrow}\hspace{2,5cm}\huge{\nearrow}\\\hspace{2,7cm}{M_X = 56}\\\hspace{1,4cm}\huge{\nearrow}\hspace{2,5cm}\huge{\searrow}\\{M_{O_2} = 32}\hspace{4cm}{8}$

$\rm \to \dfrac{n_{SO_2}}{n_{O_2}} = 3(1)$

Trong $\rm 10$ mol $\rm X$ có

$\rm n_{SO_2} + n_{O_2} = 10(2)$

Từ $\rm (1),(2)$

$\rm \to$ $\begin{cases} \rm n_{SO_2} = 7,5\\ \rm n_{O_2} =2,5  \end{cases}$

Phản ứng xảy ra

$\rm 2SO_2 + O_2 \rm \overset{t^°,\ V_2O_5}\leftrightarrows 2SO_3$

Xét tỉ lệ

$\rm \dfrac{n_{SO_2}}{2} = \dfrac{7,5}{2} > \dfrac{n_{O_2}}{1} = \dfrac{2,5}{1}$

Vậy hiệu suất theo $\rm O_2$

Gọi $\rm a$ là số mol $\rm O_2$ phản ứng

$\rm \kern1,2cm2SO_2\ \ +\ \ O_2\ \ \overset{t^°,\ V_2O_5}\leftrightarrows \ \ 2SO_3\\BĐ:\kern,25cm 7,5\kern1,22cm 2,5\kern2,3cm 0\\PỨ:\kern,4cm2a\kern1,6cma\kern2,3cm2a\\CB:7,5-2a\kern,2cm2,5-a\kern1,55cm 2a$

$\rm n_{Y} = 7,5 - 2a + 2,5 - 2a + 2a = 8$ mol

$\rm \to a = 2(mol)$

$\rm H\% = \dfrac{2}{2,5} \times 100\% = 80\%$