How many liters of water vapor can be produced if 13.3 liters of methane gas (ch4) are combusted, if all measurements are taken at the same temperature and pressure? show all of the work used to solve this problem.
ch4 (g) + 2o2 (g) yields co2 (g) + 2h2o (g)

The volume of water vapor produced is 26.6 L.

Explanation:

To calculate the volume of water vapor when number of moles changes , we use Avogadro's Law. This law states that the volume is directly proportional to the number of moles of gas if temperature and pressure remains constant.

Mathematically,

where,  

= Initial volume and number of moles

= Final volume and number of moles

For the given chemical reaction:

By Stoichiometry,

1 mole of methane reacts with 2 moles of oxygen gas to produce 1 mole oc carbon dioxide and 2 moles of water vapor.

We are given:  

Putting values in above equation, we get:

Hence, the volume of water vapor produced is 26.6 L.

the answer is 0.42 is that is an option. this question provides no info