Gas stoichiometry a sample of methane gas having a volume of 2.80 L at 25c and 1.65 atm was mixed with a sample oxygen gas having a volume 35.0L at 31c and 1.25 atm . The mixture was then ignited to form carbon dioxide and water . Calculate the volume of co2 formed at a pressure of 2.50 atm and a temperature of 125 c

2.47 L. CO2

Explanation:

First, let's establish what we need to find. The volume of CO2 formed at a pressure of 2.50 atm and a temperature of 125 C. Now, don't let this overwhelm you. To find the volume of CO2 formed, we must first find the amount of CO2 formed.

But to find the amount of CO2 formed, we have to write out the equation for this reaction. The type of reaction depicted here is combustion, where oxygen gas reacts with a substance to form carbon dioxide and water. And we also know that this equation must be balanced, according to the Law of Conservation of Matter. It is as follows:

CH4 + 2 O2 => CO2 + 2 H2O.

Now that we have the equation of the reaction, we can now find which of the two reactants is the limiting reactant. To do this, we need to convert the volumes of methane and oxygen gas to moles. Simply use the Ideal Gas Law to do this, and you get:

Methane Gas: .189 mol     |     Oxygen Gas: 1.75 mol

Using the equation, we can then see that methane gas is the limiting reactant.

We know, also due to the equation that 1 mol of CO2 is produced for every mol of methane gas. This gives us the answer of .189 mol of CO2 produced. But remember, the question asks us for the volume of CO2 produced, so we must apply the Ideal Gas Law once again, getting the answer of 2.47 L. CO2.

Note: Please remember to use significant figures during your calculations, and if you have any questions, feel free to post them down below.

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

if you need further assistance, let me know: )

~ trueboss72