24cm^3 of a mixture of methane and ethane were exploded with 90cm^3 of oxygen. After cooling to RTP the volume of gases was noted and it decreased by 32cm^3 when treated with KOH. Calculate the composition of the mixture.
To solve this problem, we need to use the concept of stoichiometry. First, let's break down the given information:
Initial volume of the mixture (V_initial) = 24 cm^3
Volume of oxygen used (V_oxygen) = 90 cm^3
Volume decrease after cooling (V_decrease) = 32 cm^3
To determine the composition of the mixture, we need to find the volume of methane (CH4) and ethane (C2H6) in the mixture.
Step 1: Determine the volume of oxygen consumed in the reaction.
From the given information, the volume of oxygen used is 90 cm^3.
Step 2: Determine the volume of carbon dioxide (CO2) produced.
According to the balanced chemical equation for the combustion of methane:
CH4 + 2O2 → CO2 + 2H2O
From the balanced equation, we can see that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide. So, the number of moles of oxygen used must be half of the number of moles of methane. Thus,
Number of moles of oxygen = 90 cm^3 * (1 mol CH4 / 2 mol O2) = 45 cm^3
Step 3: Determine the volume of carbon dioxide produced.
Since one mole of carbon dioxide is produced for every mole of methane burned, the volume of carbon dioxide (CO2) produced will be equal to the volume of methane (CH4) used. Therefore,
Volume of carbon dioxide produced = Volume of methane used = 45 cm^3
Step 4: Determine the volume of methane.
Since the volume of carbon dioxide produced is equal to the volume of methane used, the volume of methane in the mixture will be 45 cm^3.
Step 5: Determine the volume of ethane.
To find the volume of ethane, we need to subtract the volume of methane from the total initial volume of the mixture:
Volume of ethane = Initial volume of the mixture - Volume of methane = 24 cm^3 - 45 cm^3 = -21 cm^3
The negative value indicates that there is no ethane present in the mixture.
Therefore, the composition of the mixture is 100% methane and 0% ethane.
To calculate the composition of the mixture of methane and ethane, we need to use the given information and some stoichiometry.
Let's go step by step:
1. Find the initial volume of the mixture: 24 cm^3
2. Find the volume of oxygen required for complete combustion:
- For methane: 1 volume of methane reacts with 2 volumes of oxygen.
- For ethane: 1 volume of ethane reacts with 3 volumes of oxygen.
Since we have a mixture of methane and ethane, we need to calculate the individual volumes of oxygen required for complete combustion and sum them up:
Volume of oxygen for methane = (volume of methane in cm^3) * 2
Volume of oxygen for ethane = (volume of ethane in cm^3) * 3
Let's assume the volume of methane is x cm^3 and the volume of ethane is y cm^3. Therefore:
Volume of oxygen for methane = x * 2 cm^3
Volume of oxygen for ethane = y * 3 cm^3
Given that the total volume of oxygen used is 90 cm^3, we can write the equation:
x * 2 + y * 3 = 90
3. Find the volume of unreacted gas after combustion:
The volume of unreacted gas is the initial volume minus the volume of oxygen used:
Volume of unreacted gas = 24 - 90 = -66 cm^3 (negative since it is a decrease)
4. Find the volume of carbon dioxide produced:
Since all the methane and ethane have reacted, the decrease in volume after combustion (32 cm^3) must be due to the carbon dioxide produced.
Therefore, the volume of carbon dioxide produced = 32 cm^3
5. Find the volume of methane and ethane in the mixture:
Using the volume of carbon dioxide produced, we can calculate the number of moles of carbon dioxide using the ideal gas law:
PV = nRT
Here, P is the pressure at RTP (room temperature and pressure), V is the volume of carbon dioxide, n is the number of moles of carbon dioxide, R is the ideal gas constant, and T is the temperature in Kelvin.
Since we are at RTP (assumed to be 25 degrees Celsius or 298 Kelvin), the ideal gas law simplifies to:
PV = nRT
Assuming a constant pressure (P) and temperature (T), we can rearrange the equation to:
n = PV / RT
Using the known values of P, V, R, and T, we can calculate the number of moles of carbon dioxide (n).
6. Convert the volume of carbon dioxide to moles of methane:
The stoichiometric ratio between carbon dioxide and methane can be used to calculate the number of moles of methane.
From the balanced equation:
CH4 + 2O2 -> CO2 + 2H2O
We know that 1 mole of methane produces 1 mole of carbon dioxide.
Therefore, the number of moles of methane = number of moles of carbon dioxide.
Now, using the number of moles of methane and the molar volume at RTP (22.4L/mol), we can calculate the volume of methane.
7. Convert the volume of carbon dioxide to moles of ethane:
Similar to the previous step, we need to use the stoichiometric ratio between carbon dioxide and ethane to calculate the number of moles of ethane.
From the balanced equation:
C2H6 + 7/2 O2 -> 2CO2 + 3H2O
We know that 1 mole of ethane produces 2 moles of carbon dioxide.
Therefore, the number of moles of ethane = 2 * (number of moles of carbon dioxide).
Now, using the number of moles of ethane and the molar volume at RTP (22.4L/mol), we can calculate the volume of ethane.
8. Calculate the composition of the mixture:
Finally, we can use the volumes of methane and ethane to calculate their composition as a percentage.
Percentage of methane = (Volume of methane / Total volume of gases) * 100
Percentage of ethane = (Volume of ethane / Total volume of gases) * 100
Add these two percentages, and you will have the composition of the mixture.
Note: Ensure that all volumes are converted to the same units (cm^3 or L) before performing the calculations.
By following these steps, you can calculate the composition of the mixture of methane and ethane.