2. How many liters of propane gas (C3H8) will undergo complete combustion with 34.0 L of oxygen gas? ___ C3H8 (g) + ___ O2 (g → ___ CO2 (g + ___ H2O(g

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.

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