Propane burns in air to form gaseous carbon dioxide and water vapor by the equation below. If I burn 5.5 liters of propane, how many liters of water
vapor will I produce at STP?
C3H8c6 + 502(g) → 3CO2(g) + 4H2O(g)

22 L OF WATER VAPOR WILL BE PRODUCED AT STP IF 5.5 L OF PROPANE WERE BURNED.

Explanation:

Balanced Equation for the reaction:

C3H8(g) + 5O2(g) > 3CO2(g) + 4H2O(g)

From the reaction;

1 mole of propane gas reacts with 5 moles of oxygen gas to form 4 moles of water vapor.

At STP, 1 mole of a gas occupies 22.4 dm^3 of the gas

So therefore, 22.4 dm^3 of propane reacts with oxygen to produce 4 * 22.4 dm^3 of water vapor.

22.4 dm^3 of propane = 89.6 dm^3 of water vapor

If 5.5 L of propane were to be burned, how many litres of water would be produced?

22.4 L of propane produces 89.6 L of water

5.5 L of propane will produce ( 5.5 L * 89.6 L/ 22.4 L) L of water vapor

5.5 L of propane will produce 492.8 / 22.4 L of water vapor

5.5 L of propane will produce 22 L of water vapor.

So at the end of the reaction of 5.5 L of propane, 22 L of water vapor will be given off.