$\begin{array}{1 1}(a)\;\large\frac{5}{4}&(b)\;\large\frac{1}{4}\\(c)\;\large\frac{1}{2}&(d)\;\large\frac{3}{4}\end{array}$

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$SO_2+\large\frac{1}{2}$$O_2\iff SO_3$

Initial mole :$SO_2\Rightarrow 10,\large\frac{1}{2}$$O_2\Rightarrow 15$

Final mole :$SO_2\Rightarrow 10-x,\large\frac{1}{2}$$O_2\Rightarrow 15-\large\frac{x}{2},$$SO_3\Rightarrow x$

$x=8$

$\therefore SO_2: SO_3\Rightarrow 2 : 8$

$\Rightarrow 1 : 4$

Hence (b) is the correct answer.

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