Find the volume of oxygen required to burn 4.4 g of pentanol.

Let’s implement the solution:

In accordance with the condition of the problem, we write down the equation of the process:
2С5Н11ОН + 15О2 = 10СО2 + 12Н2О + Q – combustion of pentanol, carbon dioxide, water and heat are formed;

Calculations according to the formulas of substances:
M (C5H11OH) = 88 g / mol;

M (O2) = 32 g / mol.

Determine the amount of alcohol, if the mass is known:
Y (C5H11OH) = m / M = 4.4 / 88 = 0.05 mol.

Proportion:
0.05 mol (C5H11OH) – X mol (O2);

-2 mol -15 mol from here, X mol (O2) = 0.05 * 15/2 = 0.375 mol.

We find the volume of O2:
V (O2) = 0.375 * 22.4 = 8.4 L

Answer: the volume of oxygen is 8.4 liters




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