Calculate the volume, mass of carbon (IV) oxide formed by burning 56 liters of butane.

The oxidation reaction of butane with oxygen is described by the following chemical reaction equation:

2C4H10 + 13O2 = 8CO2 + 10H2O;

2 moles of gas are reacted with 13 moles of oxygen. 8 moles of carbon dioxide are synthesized.

Let’s calculate the available chemical amount of butane substance. 1 mole of ideal gas fills a volume of 22.4 liters under normal conditions.

N C3H8 = 56 / 22.4 = 2.5 mol;

Let’s calculate the volume of oxygen. To do this, multiply the amount of the substance and the standard volume of 1 mole of the gaseous substance. 1 mole of ideal gas fills a volume of 22.4 liters under normal conditions.

This will require 2.5 x 13/2 = 16.25 moles of oxygen.

V O2 = 16.25 x 22.4 = 364 liters;