Calculate the mass of water resulting from the complete combustion of 124 g of propane.

Calculate the mass of water resulting from the complete combustion of 124 g of propane. How much oxygen is needed for this?

The propane oxidation reaction is described by the following chemical reaction equation.

C3H8 + 5O2 = 3CO2 + 4H2O;

According to the coefficients of this equation, 5 oxygen molecules are required to oxidize 1 propane molecule. This synthesizes 3 molecules of carbon dioxide.

Let’s calculate the amount of propane available.

To do this, divide the available weight of the gas by the weight of 1 mole of this gas.

M C3H8 = 12 x 3 + 8 = 44 grams / mol; N C3H8 = 124/44 = 2.818 mol;

The amount of oxygen will be. N O2 = 2.818 x 5 = 14.09 mol;

Let’s calculate the gas volume. To do this, multiply the amount of substance by the volume of 1 mole of gas.

Its volume will be: V O2 = 14.09 x 22.4 = 315.6 liters;

Let’s calculate the mass of water.

N H2O = 2.818 x 4 = 11.272 mol;

M H2O = 2 + 16 = 18 grams / mol;

m H2O = 11.272 x 18 = 230 grams;