Calculate the volume of air required to burn 480kg of methane if the volume fraction of oxygen in the air is 21%.

To solve this problem, you first need to calculate how much oxygen is required to burn 480 kg of methane, and then calculate how much air contains a given volume of oxygen. But first, let’s convert kilograms to grams:

480 kg. = 480,000 g.

The methane combustion reaction can be written as an equation:

CH4 + 2 O2 = CO2 + 2 H2O

Molar mass of methane = 16 g / mol

Molar volume of oxygen = 22.4 l / mol

We make the proportion:

480,000 methane corresponds to Chl. oxygen like

16 g / mol of methane corresponds to 2 * 22.4 l / mol of oxygen

X = (480,000 * 2 * 22.4) / 16 = 1,344,000 l. oxygen.

Now let’s calculate the volume of air.

1344000 l. oxygen – 21%

V l. air – 100%

V = (1344000 * 100/21) = 6400000 l. air or 6400 m3

Answer: 6400000 liters or 6400 m3 of air