A mixture of ethyne ( C2H2 ) and methane ( CH4 ) occupied a certain volume at a total pressure of 16.8 kPa. Upon burning the sample to form CO2 and H2O, the CO2 was collected and its pressure found to be 25.6 kPa in the same volume and at the same temperature as the original mixture. What percent of the original mixture was methane?

Question

A mixture of ethyne ( C2H2 ) and methane ( CH4 ) occupied a certain volume at a total pressure of 16.8 kPa. Upon burning the sample to form CO2 and H2O, the CO2 was collected and its pressure found to be 25.6 kPa in the same volume and at the same temperature as the original mixture. What percent of the original mixture was methane?

Answer 1

The reactions are:

2C2H2(g) + 5O2(g) —> 4CO2(g) + 2H2O(L)
CH4 + 2O2 —> CO2(g) + 2H2O(L)
Let moles of C2H2 = 2x, moles CH4 = y. Then,
2xC2H2(g) + 5xO2(g) —> 4xCO2(g) + 2xH2O(L)
yCH4 + 2yO2 —> yCO2(g) + 2yH2O(L)

Original moles of gas (C2H2 + CH4) = 2x + y
Final moles of gas (CO2) = 4x + y

(4x + y) / (2x + y) = 25.6 / 16.8 = ____?
Solve for the ratio x/y
Assign the value y = 1 and solve for x.
Determine the ratio:
y / (x + y)
The above ratio times 100 is the % of methane.

Answer 2

To solve this problem, we need to understand the stoichiometry of the combustion reaction and use the ideal gas law.

First, let's write the balanced equation for the combustion of both ethyne and methane:

C2H2 + 2.5O2 --> 2CO2 + H2O
CH4 + 2O2 --> CO2 + 2H2O

From the equations, we see that for every 2 moles of ethyne, we produce 2 moles of CO2. And for every 1 mole of methane, we produce 1 mole of CO2.

Now, let's assume the initial volume of the mixture is V liters.

According to the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature (which is constant in this case).

Initially, the total pressure of the mixture is 16.8 kPa. Let's assume the number of moles of ethyne is x and the number of moles of methane is y.

Using the ideal gas law for the initial mixture:
(16.8 kPa) V = (x+y) (RT) ------- Equation 1

After burning the mixture, the CO2 pressure is 25.6 kPa. We know that the volume is the same as the initial mixture. Let's assume the number of moles of CO2 produced is z.

Using the ideal gas law for CO2:
(25.6 kPa) V = z (RT) ------- Equation 2

Since the volume (V) and temperature (T) are the same in both equations, we can divide Equation 2 by Equation 1 to eliminate V and T:

(25.6 kPa) / (16.8 kPa) = z / (x+y)

Simplifying, we get:

z = (25.6 kPa) * (x+y) / (16.8 kPa) ------- Equation 3

Now, we know that for every 2 moles of ethyne, we produce 2 moles of CO2. And, for every 1 mole of methane, we produce 1 mole of CO2.

So, the mole ratio of ethyne to methane in terms of CO2 produced is 2x : y.

We can set up another equation based on the mole ratio:

2x / y = z

Substituting the value of z from Equation 3, we get:

2x / y = (25.6 kPa) * (x+y) / (16.8 kPa)

Now we can solve this equation to find the value of x/y, which represents the mole ratio of ethyne to methane.

x/y = (25.6 kPa) * (x+y) / (16.8 kPa) * 2

Simplifying:

x/y = (25.6 kPa) * (x+y) / (33.6 kPa)

Now, we can solve this equation to find the value of x/y.

After finding the value of x/y, we can calculate the percent of the original mixture that is methane by using the relationship:

Percent methane = (y / (x+y)) * 100