AP CHEMISTRY GAS LAWS ll

Question

What volume of oxygen, at STP, is required to burn exactly 11.60 L of methane (CH4) according to the reaction below?

CH4 (g) + O2 (g) --> CO₂ (g) + H₂O (I)

Answer 1

First, we must balance the equation for the combustion of methane, CH₄.

CH₄ + 2O₂ ==> CO₂ + 2H₂O ... balanced equation

Next, we use the mol ratios between O₂ and CH₄ to determine the amount of O₂ needed.

We can also use the ratio of liters instead of moles. Thus..

11.60 L CH₄ x 2 L O₂ / 1 L CH₄ = 23.2 L of O₂

If you wanted to work with moles, and do it the long way, this is what you would have...

11.6 L CH₄ @ STP x 1 mol / 22.4 L = 0.518 mols CH₄

0.518 mols CH₄ x 2 mols O₂ / mol CH₄ = 1.036 mols O₂ needed

At STP, 1.036 mols O₂ x 22.4 L / mol = 23.2 L O₂