Final answer:
When 0.340 mol of butane is burned, 2.21 moles of oxygen are consumed, as determined by the stoichiometric relationship from the balanced combustion equation.
Step-by-step explanation:
When 0.340 mol of butane, C₄H₁₀, are burned with excess oxygen giving CO₂ and H₂O, the number of moles of oxygen consumed can be found using the balanced chemical equation for the combustion of butane:
2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(g)
From this balanced equation, we can see that every 2 moles of butane requires 13 moles of oxygen to combust completely. Using a simple stoichiometric calculation:
(0.340 mol C₄H₁₀) × (13 mol O₂ / 2 mol C₄H₁₀) = 2.21 mol O₂
Therefore, 2.21 moles of oxygen are consumed when 0.340 mol of butane is burned.