[Borek]

ok i calculated that the mass of the mixture is 0.857g

That was given, so you have not calculated it - but perhaps you expressed this information using an equation. Hard to say not seeing your work.

but how to set up the system and what is the volume of air used for

Volume and mass are required to express the density.

can I solve without the system in some way

No.

and how to use the data of 4.52 air volume. does that mean 4.52 * 0.21 for each equation

4.52*0.21 is the volume of oxygen required to burn all gases.

Assuming you have n liters of CH₄, what is the volume of oxygen required to burn it?

[Orcio_87]

@inorganiic

1. Calculate density of CH₄, H₂ and CO at normal conditions (T = 20 C, p = 1 atm).
2. If mixture composes of 100 % CH₄ it will have density you have calculated. If 100 % H₂ - some other density. If 100 % CO - yet another density.
3. Calculate how many litres of air is needed to burn 1 litre of pure CH₄, 1 litre of pure H₂, and 1 litre of pure CO.
4. You know that mixture contains only CH₄, H₂ and CO.
5. Points 2, 3 and 4 gives you system of equations (3 equations with 3 variables) to solve.

[inorganiic]

@Orcio_Dojek
I still can't solve it, I can't do the system, I calculated the densities 1.1515, 0.3290, 0.1645 and how much it takes to burn 1 dm3, 0,526 for CO and H2 and 2,107 but it doesn't work. Can you solve it I need for the entrance test. Solution is 20 percent for H2, 30 percent for CH4 and 50 percent for CO

[Orcio_87]

I calculated the densities 1.1515, 0.3290, 0.1645

What the hell? These numbers can't be correct.

Also:

how much air it takes to burn 1 dm3, 0,526 for CO and H₂ and 2,107 for CH₄

I wonder how it is possible if mixture cointains only CH₄, H₂ and CO, and for one liter it is required 4,52 liters of air

[inorganiic]

Can you solve it  because i can not almoast three days

[Orcio_87]

1. Density of CH₄, CO, H₂ and density of their mixture.
2. How much air is needed to burn 1 liter of CH₄, CO, H₂ and their mixture.
3. Mixture cointains only CH₄, CO and H₂.

Rest is only calculation.

[inorganiic]

But that is a problem i can not get 20 30 and 50 percent

[Orcio_87]

I know. Concentration of H₂ is lower than given answer.

Maybe author mistaken "normal" with "standard" (T = 273 K) conditions, I don't know.

[inorganiic]

Can you text this here how you got the answer, i would be so thankful

[Orcio_87]

@inorganiic

I must see your calculations before that.

[inorganiic]

In normal conditions we need 2,38l of air for H2 and CO from proportion, also when i made proportion i get 0,5l of O2 then for CH4 2l of O2 and 9,521l of air and then all is complicated and i can not get the right solution

[Orcio_87]

I know. My results (concentration of H₂) also differ slightly from that given by the author.

[inorganiic]

Can you write me how you got your results detailed

[inorganiic]

Or better photo them if you can

[Borek]

Can you write me how you got your results detailed

Or better photo them if you can

Per the forum rules you agreed to when registering we don't give solutions. Begging and pestering won't change that.

However, if you post how you approached the problem (and you were given multiple hints) we will gladly help you solve the problem.