1. Write the given thermochemical equations as enthalpy changes

Given:

$\mathrm{S}(g)+\frac{3}{2}\mathrm{O}_2(g) \rightarrow \mathrm{SO}_3(g)+2x\ \text{kcal}$

This means the reaction is exothermic by $2x$ kcal, so

$\Delta H_1=-2x\ \text{kcal}$

Similarly,

$\mathrm{SO}_2(g)+\frac{1}{2}\mathrm{O}_2(g) \rightarrow \mathrm{SO}_3(g)+y\ \text{kcal}$

so

$\Delta H_2=-y\ \text{kcal}$

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2. We need the heat of formation of $\mathrm{SO}_2(g)$

The formation reaction is:

$\mathrm{S}(g)+\mathrm{O}_2(g) \rightarrow \mathrm{SO}_2(g)$

Let its enthalpy change be $\Delta H_f$.

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3. Use Hess's law

We have:

$\mathrm{S}(g)+\frac{3}{2}\mathrm{O}_2(g) \rightarrow \mathrm{SO}_3(g) \qquad \Delta H=-2x$

and

$\mathrm{SO}_2(g)+\frac{1}{2}\mathrm{O}_2(g) \rightarrow \mathrm{SO}_3(g) \qquad \Delta H=-y$

Reverse the second equation:

$\mathrm{SO}_3(g) \rightarrow \mathrm{SO}_2(g)+\frac{1}{2}\mathrm{O}_2(g)$

Then

$\Delta H=+y$

Now add it to the first equation:

$\mathrm{S}(g)+\frac{3}{2}\mathrm{O}_2(g) \rightarrow \mathrm{SO}_3(g)$

$\mathrm{SO}_3(g) \rightarrow \mathrm{SO}_2(g)+\frac{1}{2}\mathrm{O}_2(g)$

Cancelling $\mathrm{SO}_3(g)$ from both sides,

$\mathrm{S}(g)+\frac{3}{2}\mathrm{O}_2(g) \rightarrow \mathrm{SO}_2(g)+\frac{1}{2}\mathrm{O}_2(g)$

Subtract $\frac{1}{2}\mathrm{O}_2(g)$ from both sides:

$\mathrm{S}(g)+\mathrm{O}_2(g) \rightarrow \mathrm{SO}_2(g)$

So this is exactly the formation reaction of $\mathrm{SO}_2(g)$.

Hence,

$\Delta H_f=(-2x)+y=y-2x\ \text{kcal}$

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4. Match with options

$\boxed{y-2x\ \text{kcal}}$

So the correct option is C.

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5. Comparison with stored correct answer

Stored correct answer: C

Derived answer: C

They agree.