Answer:
6.8 liters of propane gas will undergo a complete combustion with 34.0 L of oxygen gas.
Explanation:
Given;
combustion reaction of propane and oxygen;
C₃H₈ + O₂(g) → CO₂(g) + H₂O(g)
Balance the chemical reaction as follows;
C₃H₈ + 5O₂(g) → 3CO₂(g) + 4H₂O(g)
From the balanced equation;
1 mole of C₃H₈ ---------------- 5 moles of 5O₂(g)
x Liters of C₃H₈ ----------------- 34.0 Liters of O₂(g)
x = 34 liters / 5
x = 6.8 liters
Therefore, 6.8 liters of propane gas will undergo a complete combustion with 34.0 L of oxygen gas.
