What is the Mole Fraction of Each Gas in the Mixture?

• Chemistry
• henry3369
In summary, at 1 atm and 0°C, a 5.04 L mixture of methane (CH4) and propane(C3H8) was burned producing 20.9 g of CO2. From the given information, we can calculate that there were 0.224743 moles of the mixture present before combustion and 0.475 moles of CO2 present after the reaction. To find the mole fraction of each gas in the mixture, we can use stoichiometry and set up equations to relate the moles of each gas to the total moles of the mixture and the moles of CO2 in the product. This will give us the mole fraction of each gas, which represents the proportion of each
henry3369

Homework Statement

At 1 atm and 0° C, a 5.04 L mixture of methane (CH4) and propane(C3H8) was burned producing 20.9 g of CO2. Assume complete combustion.
1. How many moles total of methane and propane were present before combustion?

2. How many moles of carbon dioxide were present after the reactoin?

3. What was the mole fraction of each gas in the mixture?

PV = nRT

The Attempt at a Solution

1. PV = nRT
n = 0.224743 moles of mixture

2. I converted 20.9 g CO2 to moles giving:
0.475 mol of CO2

3. This is where I'm stuck.
First I balanced the combustion of the mixture:
CH4 + C3H8 + 7O2 -> 4CO2 + 6H20
Then all I have is that 0.224743 mole of the mixture. I don't know how to find the number of moles of CH4 and C3H8 each so I can find the mole fraction.

henry3369 said:

Homework Statement

At 1 atm and 0° C, a 5.04 L mixture of methane (CH4) and propane(C3H8) was burned producing 20.9 g of CO2. Assume complete combustion.
1. How many moles total of methane and propane were present before combustion?

2. How many moles of carbon dioxide were present after the reactoin?

3. What was the mole fraction of each gas in the mixture?

PV = nRT

The Attempt at a Solution

1. PV = nRT
n = 0.224743 moles of mixture

2. I converted 20.9 g CO2 to moles giving:
0.475 mol of CO2

3. This is where I'm stuck.
First I balanced the combustion of the mixture:
CH4 + C3H8 + 7O2 -> 4CO2 + 6H20
Then all I have is that 0.224743 mole of the mixture. I don't know how to find the number of moles of CH4 and C3H8 each so I can find the mole fraction.

This is where you use stoichiometry. From the moles of CO2, can you figure out how many moles C there are? Once you kno wthat you can write:

Moles C = x moles CH4 + y moles C3H8.

Where, "x" must have units of (moles C)/(moles CH4) to give you the right units. For any quantity of CH4, how many moles C are there for every mole CH4?
Same thing goes for "y" -- it must have units of (moles C)/(moles C3H8). How many moles C for every mole C3H8?

henry3369 said:
What was the mole fraction of each gas in the mixture?
When? If it was at the beginning then form 2 equations, you'll get one from total no. of moles. and the other from stoichiometry, try getting them first, as Quantum defect suggests, above

Last edited:
henry3369 said:
CH4 + C3H8 + 7O2 -> 4CO2 + 6H20

That's not combustion of the mixture. What you wrote describes - at best - what happens when you burn equimolar mixture, but your mixture is not necessarily equimolar. Each gas reacts separately, according to its own stoichiometry of combustion (hence you have two reaction equations, not one).

How does the total number of moles of CO2 in the product relate to the total number of moles of C in the reactants?

Take as a basis 0.225 moles of methane and propane, and 0.475 moles of CO2 in the product. Let x = moles of methane in the reactants and y = moles of propane in reactants. In terms of x and y, what is their relation to the 0.225 moles? In terms of x and y, what is their relation to the 0.475 moles of CO2?

Chet

1. What is mole fraction of gas in a mixture?

The mole fraction of gas in a mixture is the ratio of the number of moles of the gas to the total number of moles in the mixture.

2. How is mole fraction of gas in a mixture calculated?

To calculate the mole fraction of gas in a mixture, divide the number of moles of the gas by the total number of moles in the mixture.

3. Why is mole fraction of gas in a mixture important?

Mole fraction of gas in a mixture is important because it helps to determine the partial pressure, or the pressure exerted by a specific gas in the mixture, which is necessary for various chemical and physical processes.

4. Can the mole fraction of gas in a mixture be greater than 1?

No, the mole fraction of gas in a mixture cannot be greater than 1. It is a dimensionless quantity and is always between 0 and 1.

5. How does changing the mole fraction of gas in a mixture affect the properties of the mixture?

Changing the mole fraction of gas in a mixture can affect the properties of the mixture, such as its density and boiling point. It can also affect the rate of chemical reactions and the solubility of gases in the mixture.

• Biology and Chemistry Homework Help
Replies
5
Views
5K
• Biology and Chemistry Homework Help
Replies
2
Views
1K
• Mechanical Engineering
Replies
2
Views
608
• Biology and Chemistry Homework Help
Replies
2
Views
2K
• Biology and Chemistry Homework Help
Replies
10
Views
4K
• Biology and Chemistry Homework Help
Replies
1
Views
2K
• Biology and Chemistry Homework Help
Replies
5
Views
1K
• Thermodynamics
Replies
4
Views
1K
• Biology and Chemistry Homework Help
Replies
5
Views
1K
• Biology and Chemistry Homework Help
Replies
5
Views
3K