[Mathjr]

I've stuck to a task:

In a gas mixture (at room temperature) there is carbon monoxide, nitrogen and methane. 200 cm^3 of the mixture is heated together with 150 cm^3 oxygen. All carbon monoxide and methane are completely incinerated. The new gas mixture is cooled to the output temperature. The volume is then 220 cm^3. Lead the mixture by sodium hydroxide solution (NaOH (aq)). The volume then decreases to 80 cm^3. What does the 80 cm^3 mixture consist of?

I get the reaction formula: 2CO+CH4+3O2 => 3CO2+2H2O

Then it should cool to room temperature 20 degrees, I guess. I think I should use the gas law here but what should I use it for? To get out the pressure or amount of substance but how do I get one of these even when I have two variables now? I don't know how to proceed

[AWK]

First of all, assume that all gas volumes are given at the same temperature (not necessarily room temperature, but you have the one given). The second assumption is the complete removal of the resulting water. Finally, write CO and CH₄ oxidation reactions separately. Then you can easily calculate the volume of excess oxygen, the volume of CO₂ formed, and the volume of nitrogen in the burned mixture. Now you can do the combustion itself - a pleasant system of two equations with two unknowns.

[Mathjr]

Thank you so much for your reply. I have calculated the oxidation separately but I have just put them both in the same reaction formula:
CH4+2O2=>CO2+2H2O
2CO+O2=>2CO2
which leads to: 2CO+CH4+3O2+N=> 3CO2+2H2O+N

I still don't understand how I calculate the volume of oxygen, the volume of CO2 formed, and the volume of nitrogen in the burned mixture with this. How can I get these two unknowns? Could you show how it can be done because I have been trying to solve this for almost 3 hours now with no answers.

[Mathjr]

First of all, assume that all gas volumes are given at the same temperature (not necessarily room temperature, but you have the one given). The second assumption is the complete removal of the resulting water. Finally, write CO and CH₄ oxidation reactions separately. Then you can easily calculate the volume of excess oxygen, the volume of CO₂ formed, and the volume of nitrogen in the burned mixture. Now you can do the combustion itself - a pleasant system of two equations with two unknowns.

Thank you so much for your reply. I have calculated the oxidation separately but I have just put them both in the same reaction formula:
CH4+2O2=>CO2+2H2O
2CO+O2=>2CO2
which leads to: 2CO+CH4+3O2+N=> 3CO2+2H2O+N

I still don't understand how I calculate the volume of oxygen, the volume of CO2 formed, and the volume of nitrogen in the burned mixture with this. How can I get these two unknowns? Could you show how it can be done because I have been trying to solve this for almost 3 hours now with no answers.

[AWK]

which leads to: 2CO+CH4+3O2+N=> 3CO2+2H2O+N

This is nonsense. You assume the specific composition of the mixture and it is unknown. In addition, the nitrogen molecule is diatomic.

What does the larger volume of the mixture mean after completely burning CO and methane? What does the volume reduction of this mixture mean after washing with NaOH solution?

[sjb]

Thank you so much for your reply. I have calculated the oxidation separately but I have just put them both in the same reaction formula:
CH₄ + 2O₂  CO₂ + 2H₂O
2CO + O₂ 2CO₂
which leads to: 2CO + CH₄ + 3O₂ + N₂  3CO₂ + 2H₂O + N₂

You can only add them if the coefficients match (what is to stop you, for instance having 4CO + 2O₂  4CO₂?)

Consider what each of the bits and pieces are telling you. For instance, what does the sodium hydroxide solution do?

[Mathjr]

What does the larger volume of the mixture mean after completely burning CO and methane? What does the volume reduction of this mixture mean after washing with NaOH solution?

I don't know what it means that's why I am asking for help.

[Mathjr]

You can only add them if the coefficients match (what is to stop you, for instance having 4CO + 2O₂  4CO₂?)

Consider what each of the bits and pieces are telling you. For instance, what does the sodium hydroxide solution do?

It will react with CO2: CO2+2NaOH=>Na2CO3+H2O and the total volume I have here is 80 ml correct? I just don't know what to do with the numbers. I just feel stupid now.

[AWK]

80 ml of CO2.

 What is the volume of CO + CH4? Look at both reactions.

[Mathjr]

80 ml of CO2.

 What is the volume of CO + CH4? Look at both reactions.

130 ml

[AWK]

CH4+2O2=>CO2+2H2O
2CO+O2=>2CO2

Look at both reactions and the number of carbon atoms on both sides of each reaction and the associated volumes of carbon-containing gases.

[Mathjr]

CH4+2O2=>CO2+2H2O
2CO+O2=>2CO2

Look at both reactions and the number of carbon atoms on both sides of each reaction and the associated volumes of carbon-containing gases.

How can I calculate it when I don't know the volume of N2 and how can I use these two formulas to get the volume of these two? I mean I don't either have the substance amount or any mass.
So it would be 200-(the volume of nitrogen). I am sorry if I don't understand this, I am usually very good at solving chemistry problems but I just can't with this one.

[AWK]

The necessary information exists in these two reactions. At this point, we are looking for the total volume of CO + CH4, not the content of each of these gases.

[Mathjr]

The necessary information exists in these two reactions. At this point, we are looking for the total volume of CO + CH4, not the content of each of these gases.

I really really don't know how I do it. What does the carbon atoms show about the total volume? I just don't get it. I even asked some of my friends to look at it and they didn't know how to do it.

[AWK]

Each balanced reaction tells us about the proportions of the molecules or moles. Under the same pressure and temperature conditions, moles and gas volumes are proportional to each other.
first reaction
CH4 ... = CO2 ...
- the same number of moles - the same volumes
Second reaction
2CO ... = 2CO2 ...
- the same number of moles - the same volumes.
The total volume of CO2 is known - the conclusion: we know the volume of CH4 + CO - difficult?
This gives us the volume of nitrogen in the mixture both before the combustion of the mixture and after combustion because nitrogen does not undergo a combustion reaction. Next, we can count the excess oxygen used for combustion. This gives us the amount of oxygen consumed.
It's time to read a textbook - everything is similarly explained there.