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.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2