How to find the volume of the products in methane combustion? The combustion of methane gas, $\ce{CH4}$, gives as products $\ce{CO2}$ and $\ce{H2O}$, both in the gas phase. If $1$ liter of methane is burned in the presence of $10$ liters of $\ce{O2}$. Suppose all volumes measured under the same temperature and pressure conditions and optimal behavior for all gases.
  $$\ce{CH4 (g) + 2O2 (g) -&gt; CO2 (g) + 2H2O (g)}$$
  What is the volume of the products formed?


I know that for a gas, the volume is proportional to the amount of substance, that is, maintaining the temperature and pressure conditions, doubling the volume doubles the volume of moles.

The answer to the exercise is 3 liters.
One mole $\ce{CH4}$ reacts with 2 moles $\ce{O2}$, producing 1 mole $\ce{CO2}$ and 2 moles $\ce{H2O}$. 
So 1 liter $\ce{CH4}$ produces 1 liter $\ce{CO2}$ and 2 liter $\ce{H2O}$, if $\ce{H2O}$ is taken in the gaseous phase.