When 11.2 liters of propane were burned, gas was released. Find its volume if the product yield

When 11.2 liters of propane were burned, gas was released. Find its volume if the product yield is 85% of the theoretically possible.

Let’s execute the solution:

According to the condition of the problem, we write the equation of the process:
С3Н8 + 5О2 = 3СО2 + 4Н2О + Q – propane combustion, carbon dioxide is released.

Proportions:
1 mol of gas at normal level – 22.4 liters;

X mol (C3H8) – 11.2 liters from here, X mol (C3H8) = 1 * 11.2 / 22.4 = 0.5 mol;

0.5 mol (C3H8) – X mol (CO2);

-1 mol – 3 mol from here, X mol (CO2) = 0.5 * 3/1 = 1.5 mol.

Find the volume of the product:
V (CO2) = Y * M = 1.5 * 22.4 = 33.6 l (theoretical volume);

Y (CO2) = 0.85 * 33.6 = 28.56 L

Answer: received carbon monoxide (4) with a volume of 28.56 liters