How many liters of water vapor can be produced if 8.9 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 normal condition of the mole of a gas is 22.44L.

The volume of CH4 = 8.9L

To get the number of moles we get,

8.9 / 22.44 = 0.397 moles.

The ratio of CH4 : H2O = 1: 2

The mole of water is 0.379 × 2 = 0.794 moles.

While we convert it into liters it will be

0.794 × 22.414 = 17.80L

The answer is 17.80 liters of water vapour which will be produced.

i think a, hope it .

explanation:

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.