# Calculate the volume of air required to burn 10.4 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. This will produce 4 moles of carbon dioxide and 2 moles of water.

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

To do this, divide its weight by the weight of 1 mole of acetylene.

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

N C2H2 = 10.4 / 26 = 0.4 mol;

The required amount of oxygen will be:

N O2 = N C2H2 x 5/2 = 0.4 x 5/2 = 1 mol;

The oxygen volume will be:

V O2 = 1 x 22.4 = 22.4 liters;

The air volume will be equal to:

V air = V O2 / 0.21 = 22.4 / 0.21 = 106.7 liters;

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