How many liters of air will it take to burn 900 g methane?

1. Let’s find the amount of substance СН4.

M (CH4) = 16 g / mol.

n = 900 g: 16 g / mol = 56.25 mol.

CH4 + 2O2 = CO2 + 2H2O

For 2 mol of O2 there is 1 mol of CH4.

Substances are in quantitative ratios of 2: 1.

The amount of O2 is 2 times more than CH4.

n (O2) = 2n (CH4) = 56.25 × 2 = 112.5 mol.

Let’s find the volume of O2.

V = n Vn, where Vn is the molar volume of gas, equal to 22.4 l / mol.

V = 112.5 mol × 22.4 L / mol = 2520 L.

The air contains 21% O2.

2520 l – 21%,

X l – 100%,

X = (2520 × 100%): 21% = 12000 HP

Answer: 12,000 liters.