Chemistry help! ten points! please

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)

1.5L(1 mole/22.4L)=.066964 mole CH4


.066964 mole CH4(2 mole H2O/1 mole CH4)=.133929 mole H2O

.13929 mole H2O(22.4 L/mole)=3Liters of Vapor Water will be produced.

3L.

1 mole CH4 = 2 moles of H20
which means 1.5L CH4 = 3L H20