Methane mixed with air at 16:1 by mass

[igowithit]

Homework Statement

Methane is mixed with air at a 16:1 per mass ratio at 1atm and 298K. Calculate the mass fraction of argon, nitrogen, oxygen, and methane in the final mixture

Homework Equations

The Attempt at a Solution

First, change mass ratio of reactants to mole ratio

$$
\frac{16 g CH_4}{16 g/mol CH_4} = 1 mol CH_4
$$

$$
\frac{1 g Air}{28.97 g/mol Air} = 0.0345 mol Air
$$

or in nice round terms,

$$
29 mol CH_4
$$
$$
1 mol Air
$$

leading to

$$
29 CH_4 + 1*(0.00933Ar + 0.7809N_2 + 0.2095O_2) = aCH_4 + bCO_2 + cAr + dN_2 + eO_2 + fH_2O + gH_2 + hOH
$$

I feel like this is way too many products to reasonably allow for a timely solution. Anyways, setting up a mass balance for each of the 5 elements in the equation leads to

$$
C: 29 = a + b
$$
$$
H: 4*29 = 4a + 2f + 2g + h
$$
$$
Ar: 0.00933 = c
$$
$$
N: 2*0.7809 = 2d
$$
$$
O: 2*0.2095 = 2b + 2e + f + h
$$

we now know c and d just from these equations leaving us with 3 equations but 7 unknowns (a,b,e,f,g,h,and T). Set up $K_p$ equations based of stoichiometric equations. I'll make a guess at temperature, therefore I only have to set up 3 $K_p$ equations to go with the 3 mass balance equations I already have. However, this is where I'm stuck. When I do this, I end up with crazy complex equations that even Matlab can't solve for the coefficients.

I feel like I have entirely too many products, but I don't know how to pick and choose which ones to keep and which ones to remove. Is there any easier path to take for this problem?

 

[Chestermiller]

The problem statement doesn't say that the mixture is ignited. Are you sure it is?

 

[igowithit]

The problem statement doesn't say that the mixture is ignited. Are you sure it is?

No, the problem statement does not specify whether or not the mixture is in combustion or not. It simply states it is a mixture. I'm not sure how that changes the problem though?

EDIT: Wait, I think I see what you're saying now. I'm trying to work as a combustion problem and it's actually just an extremely simple mixture of two gases. That makes things much easier.

 

[Borek]

an extremely simple mixture of two gases

Actually a mixture of several gases, as air is a mixture itself. But I agree on the "simple" part

 

[Dr Uma Sharma]

Homework Statement

Methane is mixed with air at a 16:1 per mass ratio at 1atm and 298K. Calculate the mass fraction of argon, nitrogen, oxygen, and methane in the final mixture?

It is clearly written that it is methane mixed with air (1 atm is an indication of air) so it is 16 parts by mass of methane and one part by mass air (at 1 atm and 298K).
Use information that density of air is 1.2754 g./L (at 298K )to calculate the mass of each component of gas with its percent composition by mass .Now 1L of air is added to 16*1.2754 g of methane and this will give you 16:1 ratio .Divide each mass of individual gas (obtained above)with total mass of the sample in 1L (1.2754+16*1.2754) to get mass fraction of each component of the gas.