When 1.5 L of methane gas combust in an excess of oxygen at standard temperature and pressure, how many liters of water vapor will be produced?

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

Question

When 1.5 L of methane gas combust in an excess of oxygen at standard temperature and pressure, how many liters of water vapor will be produced? 

CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g)

anonymous · Answer

If you look at the coefficients (numbers in front of each compound) in the balanced chemical reaction, you see that there is an implied "1" in front of methane (CH4) and a "2" in front of water (H2O).  This is the ratio you need to be consistent with; if you have 1.5 L of methane as a reactant, then it follows that you will produce 3.0 L of water.