How many liters of air (air is 21% oxygen) are required for the combustion of 6.75 L of CH4? (Assume air and CH4 are at the same T and P)
Consider PV = nRT. Rewrite it as:
Everything on the right side is constant, so the n:V ratio must also be constant. That means that volume is directly proportional to the number of moles of gas.Since there is a 1:2 molar ratio between CH4 and O2.
So,from the given values, we have Volume of O2 = 2×(6.75 L CH4)
Or, 21% O2 = 13.50L of O2xL of air×100
Or, Volume of air = 13.50×10021 = 64.3 L