A 11.7-g sample of methane (CH4) reacts with 46.7 g of oxygen gas in a container fitted with a piston (at 1.00 atm and 436 K). Methane can react with oxygen to form carbon dioxide and water vapor or carbon monoxide and water vapor. After the combustion reaction is complete, the gas density at the given conditions is observed to be 0.7215 g/L. Calculate the mole fraction of methane that reacts to form carbon monoxide rather than carbon dioxide

Question

Chemistry

Queen Bee

5 August, 04:58

A 11.7-g sample of methane (CH4) reacts with 46.7 g of oxygen gas in a container fitted with a piston (at 1.00 atm and 436 K). Methane can react with oxygen to form carbon dioxide and water vapor or carbon monoxide and water vapor. After the combustion reaction is complete, the gas density at the given conditions is observed to be 0.7215 g/L. Calculate the mole fraction of methane that reacts to form carbon monoxide rather than carbon dioxide

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Jennings · Answer 1

Mole fraction of methane that reacts to form carbon monoxide rather than carbon dioxide is 0.55

Explanation:

Methane can react with oxygen to form carbon dioxide and water vapor or carbon monoxide and water vapor depending on the type of combustion that occurred. There are two types of combustion, namely; Complete and incomplete combustions.

Complete combustion occurs when methane (or any fuel) reacts with oxygen to produce carbon dioxide and water, and this occurs when the oxygen supply is sufficient for the reaction. This process produces a blue flame during combustion. But If the oxygen supply is not sufficient, an orange flame is seen, and the products is carbon monoxide, water and carbon. This is known as incomplete combustion.

The reaction of methane and oxygen to produce carbon monoxide given in the question is an example of incomplete combustion, and can be represented using the equation for the reaction below

4CH4 + 5O2 = => 2CO + 8H2O + 2C

Their respective mole ratios are given as

4 : 5==> 2 : 8 : 2

The mass of sample of Methane is 11.7g

The mass of sample of Oxygen gas is 46.7g

This means that originally, 46.7g of oxygen gas reacted with 11.7g of methane to give carbon monoxide, water and carbon in their various masses (Result not needed)

After the combustion, the gas was observed to have a density of

D = 0.7215g/L

From the question,

Pressure P=1 ATM

Temperature T = 436K

R=0.0821atmL/mol/K

M=Mass required

Recall that Density D=mass/Volume

V=m/D

And PV=nRT (ideal gas law)

n=m/M

PV = (m/M) RT

P (m/D) = (m/M) RT

(P/D) = (RT) / M

and M = (RTD) / P

M = (0.0821*436*0.7215) / 1

M=25.83g

46.7g of oxygen gas reacted with 11.7g of methane

Therefore, 25.83g of Oxygen will react with (25.83*11.7) / 46.7

=6.47g of methane.

Then, mole fraction of methane that reacts to form carbon monoxide rather than carbon dioxide = Calculated mass/original mass

=6.47/11.7

=0.55