How much air is required to burn 100 kg of zinc containing 50% of impurities?

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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