A converter is fed with 2 mol of sulphur dioxide for each 1 mol of oxygen…

Step 1. The estimate. At $1.00\ \text{bar}$ the right side is $3.39 \times 10^{4}$. With the numerator near 2, $(1 - x)^3 \approx \dfrac{2}{3.39 \times 10^{4}} = 5.9 \times 10^{-5}$, so $1 - x \approx 0.039$ and $x \approx 0.961$.

Step 2. Exact value. Put $x = 0.962$ in the left side. It gives $\dfrac{0.9254 \times 2.038}{5.5 \times 10^{-5}}$, which is close to $3.4 \times 10^{4}$. Solved to more places, $x = 0.962$ at $1.00\ \text{bar}$.

Step 3. Two bar. The right side is $6.78 \times 10^{4}$. The same method gives $x = 0.970$.

Step 4. Unconverted SO2. For each $100\ \text{mol}$ fed, the amount left is $100 \times (1 - 0.9696) = 3.04\ \text{mol}$.