If it is assumed that this reaction is taking place at STP (standard temperature and pressure) then one mole of gas occupies 22.4L
You can then calculate that 3.6L of Hydrogen contains 0.16mol
1mol/22.4L x 3.6L = 0.16mol
Every 2 moles of Hydrogen require 1 mole of oxygen to form water. Therefore half of the number of moles of oxygen are required.
0.16mol/2 = 0.08mol
Again, we know that at STP 1 mol occupies 22.4L
0.08mol x 22.4L/mol = 1.79L of O2 needed.
Also 3.6/2 = 1.8 All ideal gasses will occupy the same ammount of space at STP
