[Mathjr]

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.

First of all, I never knew that was a rule. Our mentor has never teached us that the number of moles - the same volumes. I googled this up and I got Avogadro's law, which I have never heard of before neither have my classmates and my chemistry school book doesn't even talk about it. That's a really helpful law to know thank you sir! But how do you know the total volume of CO2, it only says that 200 ml of the mixture was used with 150 ml oxygen. 130 ml of the total volume reduced after the combustion - 220 ml left. Then the mixture was led by NaOH(aq) which reduced the volume with 140 ml - 80 ml left.
I didn't understand how you got that the 80 ml was only on the CO2 also. Because I thought it was 80 ml on the products - H20 + Na2CO3. So with that also said it's 220 ml left on the total of CO2 and H2O after the combustion right? But you said before that I have to assume the removal of the resulting water but why because all the products of CH4 + 2O2 which are CO2 + 2H2O is a part of the 220 ml including the 2H2O correct?

[MNIO]

assuming you still need help, let's go back to the beginning.
  (1) you start with a mixture of CO, N2, CH4, and O2
  (2) ALL CO and ALL CH4 are incinerated. i.e.. CO --> CO2 and CH4 --> CO2 + H2O (no unreacted CO nor CH4)
  (3) we assume N2 remains unreacted
  (4) you have 200cm³ CO + N2 + CH4 and 150 cm³ O2 initially
  (5) volume of products = 220 cm³
  (6) addition of NaOH reduces volume to 80cm³

the reactions are

combustion
  x CO(g) + y CH4(g) + z N2(g) + __ O2(g) ----> (x+y) CO2(g) + 2y H2O(l) + z N2(g) + __ O2(g)

neutralization
  1 CO2(g) + 1 H2O(l) ---> 1 H2CO3(aq)
  2 NaOH(aq) + 1 H2CO3(ag) ---> 1 Na2CO3(aq) + 2 H2O

In other words, the NaOH removes the CO2 from the gaseous products and nothing else. AND the water produced is in the liquid phase (T is room T right?  For now, we're ignoring the small contribution due to saturated water vapor pressure). 

********
let's go out on a tangent for just a moment to simplify the calcs.  Take a look at this 1 reaction
  1 CH4 + 2 O2 ---> 1 CO2 + 2 H2O
those coefficients are MOLE ratios right?  We all know that by now. 

BUT in the special case of ideal gases at constant T and P, they are also VOLUME ratios
i.e
  1 mole CH4 reacts with 2 moles O2 to produce 1 mole CO2 + 2 mole H2O
AND
  1 L of CH4 reacts with 2 L of O2 to make 2 L of CO2 and 2 L of H2O

meaning... at constant T and P, we can read those coefficients as VOLUME ratio.

*********
back to the problem.

from the reaction of NaOH, we know that the final product contained
  220 cm³ - 80cm³ = 140cm³ of CO2

from the problem statement, we know we started with 150cm³ of O2

so let's update our semi-balanced reaction.
  x CO(g) + y CH4(g) + z N2(g) + 150 O2(g) ----> (x+y) CO2(g) + 2y H2O(l) + z N2(g) + __ O2(g)

where
  x + y = 140cm³
  x + y + z = 200cm³
  z = 200cm³ - 140cm³ = 60cm³
 
we know that after reaction with NaOH(aq), the only gases left are N2 and O2 so that
amount of O2 remaining unreacted is
  80cm³ - 60cm³ = 20cm³

updating again
  x CO(g) + y CH4(g) + 60 N2(g) + 150 O2(g) ----> 140 CO2(g) + 2y H2O(l) + 60 N2(g) + 20 O2(g)

now let's look at the O2..
  we started with 150cm³
  we end up with 20cm³
  therefore  130cm³ was reacted

from the reaction of CO and CH4, we can write
  2 CO + 1 O2 ---> 2 CO2
  1 CH4 + 2 O2 ---> 1 CO2 + 2 H2O

so that
  1/2 * x + 2 * y = moles O2 reacted = 130cm³
we also know
  x + y = 140

now we have 2 equations and 2 unknowns
  x/2 + 2y = 130
  x + y = 140

multiplying the first by -2
  -x + -4y = -260
  x + y = 140

adding
  -3y = -120
   y = 40
then
   x = 100

now we can write
  100 CO(g) + 40 CH4(g) + 60 N2(g) + 150 O2(g) ----> 140 CO2(g) + 2y H2O(l) + 60 N2(g) + 20 O2(g)

balancing H's
  100 CO(g) + 40 CH4(g) + 60 N2(g) + 150 O2(g) ----> 140 CO2(g) + 80 H2O(l) + 60 N2(g) + 20 O2(g)

********
now we have our balanced equation.  let's verify it makes sense
  100 + 40 + 60 = 200cm³                good
  150cm³ O2 reactant                        good
  140cm³ CO2 is stripped by NaOH      good
  60 + 20 = 80cm³ remaining gaseous products   good
and all is well

After stripping out the CO2 with NaOH, the remaining gas is
   60cm³ N2 + 20cm³ O2
or better yet
  "the 80cm³ is 0.75 mol fraction N2, 0.25 mol fraction O2"

 

[Borek]

But you said before that I have to assume the removal of the resulting water but why because all the products of CH4 + 2O2 which are CO2 + 2H2O is a part of the 220 ml including the 2H2O correct?

Water is not gaseous at STP, and the volume of the liquid is several orders of magnitude lower, so can be neglected.

[Mathjr]

from the reaction of CO and CH4, we can write
  2 CO + 1 O2 ---> 2 CO2
  1 CH4 + 2 O2 ---> 1 CO2 + 2 H2O

so that
  1/2 * x + 2 * y = moles O2 reacted = 130cm³

Now I understood everything thank you sir! But this part here got me a little bit confused, why 1/2 and not just 1.

[Borek]

What is the molar ratio CO/O₂ in the reaction?

[Mathjr]

What is the molar ratio CO/O₂ in the reaction?

Don't you mean O2/CO? Because that's the only way I see him get that answer. Did he just:
(2 CO)/2 + (1 O2)/2 ---> (2 CO2)/2
1 CO + 1/2 O2 ---> 1 CO2
Could you briefly explain how he got those numbers and why he multiplied it with x and y? Is it about Avogadro's law?

[Borek]

If you don't understand how, don't worry, this is not an approach that is worth using IMHO.

If x is number of moles CO, how many moles of O₂ will react with it?

[Mathjr]

If you don't understand how, don't worry, this is not an approach that is worth using IMHO.

If x is number of moles CO, how many moles of O₂ will react with it?

There is only 1 mole of O2 so 2 mole? I really don't get it how can that be done.

What is the molar ratio CO/O₂ in the reaction?

The molar ratio is 2:1 because for every 1 mole O used, 2 moles of CO are formed

[Mathjr]

If you don't understand how, don't worry, this is not an approach that is worth using IMHO.

If x is number of moles CO, how many moles of O₂ will react with it?

There is only 1 mole of O2 so 2 mole? I really don't get it how can that be done. I don't also get why it is multiplied with x and y, the volumes.

What is the molar ratio CO/O₂ in the reaction?

The molar ratio is 2:1 because for every 1 mole O used, 2 moles of CO are formed. Also there is one thing I still wonder. Why is this wrong to say: N2+2CO+CH4+3O2=>3CO2+2H2O+O2+N2
My friends used this and they still got 60 ml N and 20 ml O. But that's not so important, the important part is why is that wrong to say? I have to send the answer to my teacher this day and I would appreciate if you could answer asap because these are the key things I need to understand rn.

[MNIO]

<sigh>
I wrote
  2 CO + 1 O2 ---> 2 CO2
  1 CH4 + 2 O2 ---> 1 CO2 + 2 H2O

so that
  1/2 * x + 2 * y = cm³ O2 reacted = 130cm3

*********
look at the stoichiometry
  for every 2 mL of CO reacted, 1 mL of O2 reacted
  for every 1 mL of CH4 reacted, 2 mL O2 reacted
so that
  mL O2 reacted = 1/2 * mL CO + 2 * mL CH4

or if you prefer
   mL O2 reacted = x mL CO * (1 mL O2 / 2 mL CO) = 1/2 * x
   mL O2 reacted = y mL CH4 * (2 mL O2 / 1 mL CH4) = 2y
then
  total mL O2 reacted = 1/2 * x + 2y = 130mL

[Mathjr]

<sigh>
I wrote
  2 CO + 1 O2 ---> 2 CO2
  1 CH4 + 2 O2 ---> 1 CO2 + 2 H2O

so that
  1/2 * x + 2 * y = cm³ O2 reacted = 130cm3

*********
look at the stoichiometry
  for every 2 mL of CO reacted, 1 mL of O2 reacted
  for every 1 mL of CH4 reacted, 2 mL O2 reacted
so that
  mL O2 reacted = 1/2 * mL CO + 2 * mL CH4

or if you prefer
   mL O2 reacted = x mL CO * (1 mL O2 / 2 mL CO) = 1/2 * x
   mL O2 reacted = y mL CH4 * (2 mL O2 / 1 mL CH4) = 2y
then
  total mL O2 reacted = 1/2 * x + 2y = 130mL

Okey and one last question before I stop. AWK said it's wrong to say that the reaction formula is:

CH4 + 2O2 => CO2 + 2H2O
2CO + O2 => 2CO2
N2 + 2CO + CH4 + 4O2 => 3CO2 + 2H2O + O2 + N2

With the argument that I can only add them if the coefficients match. Instead, It should like this like you did:

x CO(g) + y CH4(g) + z N2(g) + 150O2(g) ----> (x+y) CO2(g) + 2y H2O(l) + zN2(g) + 20O2(g)
Why does this work. I don't understand why the reaction formula can be written like this because it's not even balanced and why do we have to take 2 moles times the volume of water and not 2 moles times CO? I don't understand this. I know that you have right but no one of my classmates have solved it like this because they used this formula: N2 + 2CO + CH4 + 4O2 => 3CO2 + 2H2O + O2 + N2

[MNIO]

If you don't understand how, don't worry, this is not an approach that is worth using IMHO.

If x is number of moles CO, how many moles of O₂ will react with it?

What do you mean not an approach worth using?  This is the standard approach to solving this particular problem variation.  Which is a variation of balancing an equation.  Which most of us have seen and solved dozens if not hundreds of times over our time in gen chem, intro to ChE and TA'ing classes.  How would you do it differently?  It's not a difficult problem once you (1) write the unbalanced equation, (2) realize you can substitute volume for moles, (3) recognize that NaOH removes CO2 and H2O is (l). right?

[Mathjr]

If you don't understand how, don't worry, this is not an approach that is worth using IMHO.

If x is number of moles CO, how many moles of O₂ will react with it?

What do you mean not an approach worth using?  This is the standard approach to solving this particular problem variation.  Which is a variation of balancing an equation.  Which most of us have seen and solved dozens if not hundreds of times over our time in gen chem, intro to ChE and TA'ing classes.  How would you do it differently?  It's not a difficult problem once you (1) write the unbalanced equation, (2) realize you can substitute volume for moles, (3) recognize that NaOH removes CO2 and H2O is (l). right?

Can you please reply I have spent this whole day on this problem.

[MNIO]

Mathjr

The majority of this problem is simply an exercise in writing and balancing equations.  That's it.  You need to predict the products, recognize that H2O is liquid and that NaOH removes CO2... And to simplify things, you need to know that at constant T and P, you can substitute volume for moles in that balanced equation.

**********
from the problem statement, we assume
  (1) ALL CO ---> CO2
  (2) ALL CH4 ---> CO2 + H2O
  (3) N2 remains unreacted
  (4) Some O2 remains unreacted

so that
  __CO(g) + __ CH4(g) + __ N2(g) + __ O2(g) ----> __ CO2(g) + __ H2O(l) + __ N2(g) + __ O2(g)

now let's balance this thing

*******
starting with what we know
  (1) we know CO + CH4 + N2 = 200
        but we don't know mL of each.. just the total
  (2) so let's use x, y, and z for those values (coefficients)
and write
  x CO(g) + y CH4(g) + z N2(g) + __ O2(g) ----> __ CO2(g) + __ H2O(l) + __ N2(g) + __ O2(g)

next let's find the coefficients of CO2 and H2O in terms of x and y
    balancing C's
        we know that from this reaction
               1 CO + 1/2 O2 ---> 1 CO2
            for every 1 CO we consume, we produce 1 CO2
            therefore for every x CO we consume, we produce x CO2
        from this reaction
               1 CH4 + 2 O2 ---> 1 CO2 + 2 H2O
            for every 1 CH4 we consume, we produce 1 CO2
            so for every y CH4 we consume, we produce y CO2
        so overall
            for every x O2 + y CH4, we produce (x CO2 + y CO2) = (x + y) CO2
        which is the same thing as writing
            x CO2 + y CH4 ---> (x + y) CO2

    balancing H's
        from this reaction
             1 CH4 + 2 O2 ---> 1 CO2 + 2 H2O
        for every 1 CH4 we consume we produce 2 H2O
        so for every y CH4 we consume we produce 2y H2O

now we have
  x CO(g) + y CH4(g) + z N2(g) + __ O2(g) ----> (x+y) CO2(g) + 2y H2O(l) + __ N2(g) + __ O2(g)

continuing balancing, we assume N2 is unreacted so
  x CO(g) + y CH4(g) + z N2(g) + __ O2(g) ----> (x+y) CO2(g) + 2y H2O(l) + z N2(g) + __ O2(g)

next, we know O2 initial, but we don't know amount of unreacted O2
so let's write this
  x CO(g) + y CH4(g) + z N2(g) + 150 O2(g) ----> (x+y) CO2(g) + 2y H2O(l) + z N2(g) + __ O2(g)

next, let's work on CO2.
   we know volume of CO2(g) + N2(g) + O2(g) = 220
   we know volume of N2(g) + O2(g) = 80
so
   volume CO2(g) = 220 - 80 = 140

updating our partially "balanced" equation
  x CO(g) + y CH4(g) + z N2(g) + 150 O2(g) ----> 140 CO2(g) + 2y H2O(l) + z N2(g) + __ O2(g)

now we know
  x + y + z = 200
  x + y = 140
therefore
  z = 200 - 140 = 60

now we have this
  x CO(g) + y CH4(g) + 60 N2(g) + 150 O2(g) ----> 140 CO2(g) + 2y H2O(l) + 60 N2(g) + __ O2(g)

and we know that
  N2 + O2 = 80   right?  the gas remaining after CO2 was removed?
so
  O2 = 80 - 60 = 20

and we can write
  x CO(g) + y CH4(g) + 60 N2(g) + 150 O2(g) ----> 140 CO2(g) + 2y H2O(l) + 60 N2(g) + 20 O2(g)

and we can stop there are write
   remaining gas = 0.75 mole fraction N2 and 0.25 mole fraction O2

********
continuing on.. if you want

we can see that 150 - 20 = 130 mL of O2 reacted with the x CO + y CH4

from this reaction
      2 CO + 1 O2 --> 2 CO2
   for every 1 CO reacted, we consume 1/2 O2
   so for every x CO reacted, we consume 1/2x O2

from this reaction
      1 CH4 + 2 O2 ---> 1 CO2 + 2 H2O
   for every 1 CH4 reacted, we consume 2 O2
   so for every y CH4 reacted, we consume 2y O2

so that
  x CO + y CH4 requires 1/2x + 2y total O2

since we know O2 consumed = 130, we can write
  1/2x + 2y = 130

and we know that
  x + y + z = 200
  z = 60
  x + y = 200 - 60 = 140

now we have 2 equations with 2 unknowns
  1/2 x + 2y = 130
  x + y = 140

solve for x and y however you like... the result is x = 100, y = 40, 2y = 80

and now we can write...
  100 CO(g) + 40 CH4(g) + 60 N2(g) + 150 O2(g) ----> 140 CO2(g) + 80 H2O(l) + 60 N2(g) + 20 O2(g)

*******
Again, this entire problem is writing and "balancing" the equation.  Practice it a few times and recognize the variation of this problem in the future.  You'll probably see it again.   

[Borek]

What do you mean not an approach worth using?

IMHO in this case it is much better to solve the problem with two separate balanced equations. Not because your approach is invalid, but because using single equation when someone is struggling with basic concepts has a negative pedagogical value, instead of helping produces more confusion. What you call "balancing an equation" is not what books and teachers teach, it is a related, but separate concept (not a "balanced equation" but "equation that represents stoichiometry of a chemical process"). Once people have the basics right and do such problems easily, then trying a combined stoichiometry equation can be valuable. Not before.