What volume of oxygen is required for the complete combustion of 1 mol of cycloalkane

What volume of oxygen is required for the complete combustion of 1 mol of cycloalkane, the relative vapor density of which for oxygen is 1.75?

Given:
Cycloalkane – CnH2n
n (CnH2n) = 1 mol
D O2 (CnH2n) = 1.75

Find:
V (O2) -?

Solution:
1) M (O2) = Mr (O2) = Ar (O) * 2 = 16 * 2 = 32 g / mol;
2) M (CnH2n) = D O2 (CnH2n) * M (O2) = 1.75 * 32 = 56 g / mol;
3) M (CnH2n) = Mr (CnH2n) = Ar (C) * n + Ar (H) * 2n = 12 * n + 1 * 2n = (14n) g / mol;
4) 14n = 56;
n = 4;
Unknown cycloalkane – C4H8 – cyclobutane;
5) C4H8 + 6O2 => 4CO2 + 4H2O;
6) n (O2) = n (C4H8) * 6 = 1 * 6 = 6 mol;
7) V (O2) = n (O2) * Vm = 6 * 22.4 = 134.4 liters.

Answer: The O2 volume is 134.4 liters.