Please solve this question

This question guys

You can confidently say 1 mole of CnH2n will make n moles of CO2 and n moles of H2O, so you get the unbalanced equation
CnH2n (g) + O2 (g) —> nCO2 (g) + nH2O (l)
Counting the number of oxygens on the RHS: 2 x n + 1 x n = 3n, so there must be 1.5n moles of O2 on the LHS, e.g
CnH2n (g) + 1.5nO2 (g) —> nCO2 (g) + nH2O (l)
Since the moles of gas and the volume occupied by a gas are proportional, you can treat the volumes like moles. If you burn 100 cm^3 of CnH2n, you need 150n cm^3 of O2 (1.5n x 100 cm^3 = 150n cm^3). That is to say the volume of oxygen left is (800 - 150n) cm^3.
But the leftover oxygen isn’t the only gas in the mixture. The CO2 made is also a gas and there should be 100n cm^3 of it. That means the total volume of gas left is (800 - 150n) cm^3 + 100n cm^3 = (800 - 50n) cm^3
But we were told that the total volume of gas left is 600 cm^3, i.e
600 = 800 - 50n => 50n = 200, so n = 4
Therefore the gas is C4H8.