How much oxygen is required to burn 11.4 kg of octane.

Let’s execute the solution. Let’s compose the reaction equation:
2С8Н18 + 25О2 = 16СО2 + 18Н2О – octane combustion reaction, carbon dioxide and water are released;
Let’s determine the molecular weights of substances:
M (C8H18) = 12 * 8 + 1 * 18 = 114 g / mol;
Let’s calculate the number of moles of octane, if the mass is known:
Y (C8H18) = m / M = 11.4 / 114 = 0.1 mol;
We find the value of the number of mol O2:
0.1 mol (C8H18) – X mol (O2);
-2 mol -25 mol from here, X mol (O2) = 0.1 * 25/2 = 1.25 mol;
Let’s calculate the volume of oxygen using Avogadro’s law:
V (O2) = 1.25 * 22.4 = 28 liters. = 28 m3.
Answer: in the reaction of octane combustion, oxygen with a volume of 28 m3 is required.