How many liters of CO2 are generated by combustion of: a) 224 g of pentanol b) 24 g of octene?

| March 12, 2021 | education

Let’s execute the solution:

2С5Н11ОН + 15О2 = 10СО2 + 12Н2О + Q – combustion of pentanol, carbon dioxide is released;
С8Н16 + 12О2 = 8СО2 + 8Н2О + Q – octene burns with the release of carbon monoxide (4).
Calculations:
M (C5H11OH) = 88 g / mol.

M (C8H16) = 112 g / mol.

Let us determine the amount of starting substances, if the mass is known:
Y (C5H11OH) = m / M = 224/88 = 2.5 mol.

Y (C8H18) = m / M = 24/112 = 0.2 mol.

5. Proportions:

2.5 mol (C5H11OH) – X mol (CO2);

-2 mol – 10 mol from here, X mol (CO2) = 2.5 * 10/2 = 12.5 mol.

0.2 mol (C8H16) – X mol (CO2);

-1 mol – 8 mol from here, X mol (CO2) = 0.2 * 8/1 = 1.6 mol.

Find the volume of the product:
V (CO2) = 12.5 * 22.4 = 280 l;

V (CO2) = 1.6 * 22.4 = 35.8 liters.

Answer: when pentanol burns, carbon dioxide is released in a volume of 280 liters, octene burns out, carbon monoxide is formed in a volume of 35.8 liters.



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