A mixture of ethyne gas (C2H2) and methane gas (CH4) occupied a certain volume at a total pressure of 16.8 kPa. When the sample burned, the products were CO2 gas and H2O vapor. The CO2 was collected and its pressure found to be 25.2 kPa in the same volume and same temperature as the original mixture. What percentage of the original mixture was methane?
Check my thinking on this.
The gas starts as 16.8 kPa Ptotal.
It ends up as 25.2 kPa Ptotal.
Look at the equations.
C2H2 + 5/2 O2 = 2CO2 + H2O
CH4 + 2O2 ==> CO2 + 2H2O
You can see that 1 mol C2H2 produces 2 moles CO2 while 1 mol CH4 produces 1 mol CO2. IF WE HAD 100% CH4, we would expect to obtain 16.8 kPa CO2. The difference (25.2-16.8 = 8.4) must be due to the C2H2 present. Then 16.8 = Ptotal and 8.4 = PC2H2 and 16.8-8.4 = 8.4 PCH4. So we are starting with 8.4 kPa CH4 and 8.4 kPa C2H2 or 50% each.
wrong, answer is that the methane makes up 46% of the gases being burned. (I have the chem textbook that the Q comes from).
To find the percentage of methane in the original mixture, we need to compare the partial pressures of methane and ethyne in the mixture.
1. Let's assume the partial pressure of methane is P(CH4) and the partial pressure of ethyne is P(C2H2).
2. According to Dalton's Law of Partial Pressures, the total pressure of the mixture is the sum of the partial pressures of the individual gases. Therefore, we have:
P(CH4) + P(C2H2) = 16.8 kPa
3. When the mixture burns, methane is converted to CO2. So, the partial pressure of CO2 (P(CO2)) after the burning process would be equal to the partial pressure of methane (P(CH4)) in the original mixture.
4. We are given that the pressure of CO2 is 25.2 kPa. Therefore, P(CO2) = 25.2 kPa.
5. Using the information from steps 3 and 4, we have:
P(CH4) = 25.2 kPa
6. Substituting the value of P(CH4) from step 5 into the equation from step 2, we can solve for P(C2H2):
25.2 kPa + P(C2H2) = 16.8 kPa
P(C2H2) = 16.8 kPa - 25.2 kPa
P(C2H2) = -8.4 kPa
7. We know that the pressure of a gas cannot be negative, so we can conclude that there was no ethyne in the original mixture.
8. Now, we can find the percentage of methane in the original mixture by comparing the partial pressure of methane to the total pressure of the mixture:
Percentage of methane = (P(CH4) / Total pressure) x 100
Percentage of methane = (25.2 kPa / 16.8 kPa) x 100
Percentage of methane = 150%
Therefore, the original mixture was 150% methane (we obtained a higher percentage due to the negative pressure value for ethyne, indicating its absence).
To determine the percentage of methane in the original mixture, we need to calculate the moles of CO2 produced from the combustion reaction.
First, let's assign some variables:
Let n1 be the moles of ethyne (C2H2) in the original mixture.
Let n2 be the moles of methane (CH4) in the original mixture.
From the balanced combustion equation:
C2H2 + 2.5 O2 → 2 CO2 + H2O
1 mole of ethyne produces 2 moles of CO2.
And since there are no other sources of CO2, the moles of CO2 produced should be equal to n1 × 2.
Given that the pressure of CO2 (P2) is 25.2 kPa and the total pressure of the mixture (P1) is 16.8 kPa, we can use Dalton's Law of partial pressures:
P2 = X2 × P1
where X2 is the mole fraction of CO2 in the mixture.
Rearranging the equation, we get:
X2 = P2 / P1
Using this expression, we can find the mole fraction of CO2.
Since there are only two components in the original mixture (ethyne and methane), we can write:
X2 = n2 / (n1 + n2)
Substituting the expression for X2 in terms of P2 and P1, we have:
P2 / P1 = n2 / (n1 + n2)
Now we can substitute the given values:
P2 = 25.2 kPa
P1 = 16.8 kPa
Solving for n2, we have:
25.2 kPa / 16.8 kPa = n2 / (n1 + n2)
Cross-multiplying, we get:
25.2 kPa * (n1 + n2) = 16.8 kPa * n2
25.2 n1 + 25.2 n2 = 16.8 n2
25.2 n1 = 16.8 n2
n1 / n2 = 16.8 / 25.2
n1 / n2 = 2 / 3
Therefore, the ratio of moles of ethyne (C2H2) to methane (CH4) in the original mixture is 2:3. This means that for every 2 moles of ethyne, there are 3 moles of methane.
To find the percentage of methane, we calculate:
Percentage of methane = (n2 / (n1 + n2)) * 100
Percentage of methane = (3 / (2 + 3)) * 100
Percentage of methane = 3/5 * 100
Percentage of methane = 60%
Hence, the percentage of methane in the original mixture is 60%.