How many liters of propane (C3H8) are needed to react with 15.2 liters of oxygen to produce water and carbon dioxide

3.04 liters of propane (C₃H₈) are needed to react with 15.2 liters of oxygen to produce water and carbon dioxide

Explanation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

The mole ratio between the propane and oxygen in the equation is 1 : 5 as can be seen in the equation.

If the amount of O₂ reacting is 15.2 L then the amount of propane required is;

1 : 5

X : 15.2

We cross multiply the ration

5x = 15.2 * 1

X = 15.2/5

X = 3.04

= 3.04 L