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When a mixture of 10 moles of $$\text{SO}_2$$ and 16 moles of $$\text{O}_2$$ were passed over a catalyst, 8 moles of $$\text{SO}_3$$ were formed at equilibrium. The number of moles of $$\text{SO}_2$$ and $$\text{O}_2$$ remaining unreacted were
The reaction is $$2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3$$. Forming 8 mol of $$\text{SO}_3$$ uses 8 mol of $$\text{SO}_2$$ and 4 mol of $$\text{O}_2$$. So $$10 - 8 = 2$$ mol of $$\text{SO}_2$$ and $$16 - 4 = 12$$ mol of $$\text{O}_2$$ remain.
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