How much air is required to burn 100 kg of zinc containing 50% of impurities?
July 25, 2021 | education
| The reaction of zinc with oxygen takes place according to the following chemical reaction equation:
Zn + ½ O2 = ZnO;
For oxidation of 1 mol of metal, 0.5 mol of oxygen is required.
Let’s find the amount of metal.
The mass of pure metal is 100 x 0.5 = 50 kilograms;
M Zn = 65 grams / mol;
N Zn = 50,000/65 = 769.23 mol;
To burn this amount of metal, 769.23 / 2 = 384.6 mol of oxygen is required.
Let’s find its volume.
One mole of gas under normal conditions fills a volume of 22.4 liters.
The oxygen volume will be equal to:
V О2 = 384.6 x 22.4 = 8 615 liters;
The volumetric concentration of oxygen in the air is 20.95%.
The required air volume will be:
V air = 8 615 / 0.2095 = 41122 liters = 41.122 m3;
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