How much oxygen is required to burn 22 liters of methane?

To solve this problem, we write down given: V (CH4) = 22 liters. Methane was burned in oxygen.

Find: the amount of oxygen needed to burn methane.

Solution:

Let’s write down the reaction equation.

CH4 + O2 = CO2 + H2O

Let’s arrange the coefficients.

CH4 + 2O2 = CO2 + 2H2O

Above methane, we write 22 liters, and under methane, we write a constant molar volume, which is 22.4 liters / mol.

Methane entered 1 mol, therefore 1 mol * 22.4 L / mol = 22.4 L

Above oxygen we write x l, and under oxygen a constant molar volume, which is 22.4 l / mol.

Since 2 mol of oxygen entered, 2 mol * 22.4 l / mol = 44.8 l

Let’s compose and solve the proportion.

x = 22 * ​​44.8 / 22.4 = 44 l

Answer: 44 l