How much air is required for the complete combustion of 108 g of acetylene?

The oxidation reaction of acetylene with oxygen is described by the following chemical reaction equation:

2C2H2 + 5O2 = 4CO2 + 2H2O;

2 moles of acetylene are reacted with 5 moles of oxygen.

Let’s calculate the available chemical amount of the acetylene substance.

M C2H2 = 12 x 2 + 1 x 2 = 26 grams / mol;

N C2H2 = 108/26 = 4.154 mol;

To burn such an amount of matter, you need to take 4.154 x 5/2 = 10.385 mol of oxygen.

Let’s calculate its volume. To do this, multiply the amount of the substance and the standard volume of 1 mole of the gaseous substance. 1 mole of ideal gas fills a volume of 22.4 liters under normal conditions.

V O2 = 10.385 x 22.4 = 232.6 liters;

The oxygen content in the air is 21%. The required air volume will be:

V air = 232.6 / 0.21 = 1 108 liters;