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