How much oxygen is required for the complete combustion of 26.5 g of ethylbenzene?

How much oxygen is required for the complete combustion of 26.5 g of ethylbenzene? What are all the possible isomers of a given hydrocarbon and name them according to the systematic nomenclature?

In accordance with the data of the problem, we write the equation for the combustion of ethylbenzene:
2С8Н10 = 21О2 = 16СО2 + 10Н2О + Q – heat, carbon dioxide, water is released;

We make calculations:
M (C8H10) = 106 g / mol;

M (O2) = 32 g / mol;

Y (C8H10) = m / M = 26.5 / 106 = 0.25 mol.

Proportion:
0.25 mol (C8H10) – X mol (O2);

-2 mol – 21 mol from here, X mol (O2) = 0.25 * 21/2 = 2.625 mol.

We find the volume of O2:
V (O2) = 2.625 * 22.4 = 58.8 L

Isomers:
C6H5 – (CH3) – (CH3) – 1,3 dimethyl benzene (ortho-xylene);

C6H5 – (CH3) – (CH3) – 1,4 dimethyl benzene (para-xylene).

Answer: you need oxygen with a volume of 58.8 liters