For the reaction A(g) ⇌ B(g) at 495K , Δ _ r G ^ ° = - 9 . 478kJ...

$\Delta G ° = - {RTℓnK}_{eq}$

Given $\Delta G ° = - 9 . 478KJ / mole$

$T = 495KR = 8 . 314J{mol}^{- 1}$

So $- 9 . 478 \times {10}^{3} = - 495 \times 8 . 314 \times ℓ{nK}_{eq}$

$ℓ{nK}_{eq} = 2 . 303$

$= ℓn10$

So ${K}_{eq} = 10$

Now $\begin{cases} & A\left(g\right) & ⇌ & B\left(g\right) \\ t = 0 & 22 & & 0 \\ t = 0 & 22 - x & & x\end{cases}$

${K}_{eq} = \dfrac{\left[B\right]}{\left[C\right]} = \dfrac{x}{22 - x} = 10$

or $x = 20$

So millimoles of $B = 20$