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.

ljjjljlj kjlk,ljym jlyu; jm; l mumytj

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error