How many liters of carbon monoxide (IV) will be released during combustion 63g. propane.

Let’s write down the solution sequentially:
1. We compose the reaction equation:
С3Н8 + 5О2 = 3СО2 + 4Н2О – propane combustion occurs with the release of carbon dioxide;
2. Determine the amount of propane if its mass: m (C3H8) = 63g, M (C3H8) = 44 g / mol;
Y (C3H8) = 63/44 = 1.43 mol.
1. Let’s make a proportion, taking into account the data on the reaction equation:
1.43 mol (C3H8) – X mol (CO2);
-1 mol -3 mol from here,
X mol (CO2) = 3 * 1.43 / 1 = 4.29 mol.
5. Let’s calculate the volume of CO2: V (CO2) = 4.29 * 22.4 = 96.09 liters.
Answer: during the combustion of propane, 96.09 liters were released.