Question

A 2.50-l flask contains a mixture of methane (ch4) and propane (c3h8) at a pressure of 1.45 atm and 20°c. when this gas mixture is then burned in excess oxygen, 8.60 g of carbon dioxide is formed. (the other product is water.) what is the mole fraction of methane in the original gas mixture?

Answer 1

Answer:- Mole fraction of methane in the original gas mixture is 0.854.

Solution:- From given volume, pressure and temperature, we could calculate the total moles of the gaseous mixture of methane and propane using ideal gas law as:

PV = nRT

`n=(PV)/(RT)`

V = 2.50 L

P = 1.45 atm

T = 20 + 273 = 293 K

Let's plug in the values in the equation:

`n=((1.45*2.50)/(0.0821*293))`

n = 0.151

Let's say the solution has X moles of methane. Then moles of propane would be = (0.151 - X)

The combustion equations of methane and propane are:

`CH_4+2O_2\rightarrow CO_2+2H_2O`

`C_3H_8+5O_2\rightarrow 3CO_2+4H_2O`

From methane balanced equation, there is 1:1 mol ratio between methane and carbon dioxide. So, X moles of methane would produce X moles of carbon dioxide.

From balanced equation of propane, there is 1:3 mol ratio between propane and carbon dioxide. So, (0.151 - X) moles of propane would give 3(0.151 - X) moles of carbon dioxide.

So, total moles of carbon dioxide that we would get from methane and propane combustion are:

X + 3(0.151 - X)

From given data, 8.60 g of carbon dioxide are formed by the combustion of gas mixture.

moles of Carbon dioxide =
`8.60g((1mol)/(44g))`

moles of carbon dioxide = 0.195 mol

Hence, 0.195 = X + 3(0.151 - X)

Let's solve this for X as:

0.195 = X + 0.453 - 3X

0.195 = 0.453 - 2X

2X = 0.453 - 0.195

2X = 0.258

`X=(0.258)/(2)`

X = 0.129

So, there are 0.129 moles of methane in the mixture.

moles of propane = 0.151 - 0.129 = 0.022

mole fraction of methane =
`(moles of Methane)/(total moles)`

mole fraction of methane =
`(0.129)/(0.151)`

mole fraction of methane = 0.854

Hence, the mole fraction of methane gas in the original gas mixture is 0.854.