How many liters of air is consumed to burn 53 g of ammonia (NH3)?
| July 7, 2021 | education
The reaction of ammonia with oxygen is described by the following equation:
4NH3 + 3O2 = 2N2 + 6H20;
Combustion of 4 moles of ammonia requires 3 moles of oxygen.
The molar mass of ammonia is 14 x 3 = 17 grams / mol.
The number of moles of the substance in 53 grams of ammonia is 53/17 = 3.1176 mol.
The required amount of oxygen will be 3.1176 * 3/4 = 2.3382 mol.
Under normal conditions, 1 mole of ideal gas takes up a volume of 22.4 liters.
2.3382 moles of oxygen will occupy a volume of 2.3382 x 22.4 = 52.3757 liters.
The oxygen content in the air by volume is 20.95%.
Accordingly, the required air volume will be 52.3757 / 0.2095 = 250 liters.
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