1) 30cm3 of methane and ethane mixed with 100cm3 of O2 at RTP. Explosion occurs, on cooling 61.5cm3 remained. Find the percentage by volume of each gas in the mixture

2)32cm3 of a mix of CO2, CH4 and H2 mix with 50cm3 of O2 and exploded after cooling to RTP, volume noted is reduced by 22cm3 when exposed to KOH solution, 16cm3 of O2 is in excess. Calculate the composition of the mixture.

3) 24cm3 of a mixture of CH$ and C2H6 exploded with 90cm3 of O2 after cooling to RTP, volume was noted which was decreased by 32cm3 when treated with KOH solution. Calculate the composition of the mixture.

4) 195cm3 of hydrogen, carbon monoxide and nitrogen mixed with 150cm3 of oxygen exploded. Final volume after cooling is 165cm3. Final volume after absorbtion with KOH is 75cm3. Find the percentage by volume of each of the gases present.

3 is the easiest. The others are the same kind but are a little more complicated. This will get you stated on #3.

CH4 + 2O2 ==> CO2 + H2O
2C2H6 + 7O2 ==> 4CO2 + 6H2O

If O2 is 90 cc initially and excess O2 was 32 cc, then 90-32 = 58 cc O2 used.
Let x = volume CH4
and (24-x) = volume C2H6

volume O2 used for CH4 is x*(2 mol O2/1 mol CH4) = 2x
volume O2 used for C2H6 is (24-x)*7/2 = 168-7x/2.
The sum of O2 used by CH4 and C2H6 is
2x + 168-7x/2 = 58
Solve for x and I get about 17cc O2 used for CH4 and about 7 cc for O2 used for combustion of C2H6.

Note that these are approximations. You need to go through the math and obtain better accuracy. Then use cc O2 used for each to calculate volume CH4 and volume C2H6 and from there you can find percentage.

1) Well, it looks like we have a mixture that went "BOOM!" and we need to figure out the percentage by volume of each gas. Let's get cracking!

First, let's look at the initial volume of the gases:
- We have 30cm3 of methane, let's call it "M" for short.
- We also have 30cm3 of ethane, let's call it "E" for easy.

Now, let's see what happened after the explosion:
- After the explosion and cooling, we only have 61.5cm3 of gas left.
- Since we started with 30cm3 of M and 30cm3 of E, that means we lost 30cm3 of gas.

So, we can conclude that the remaining 61.5cm3 must be made up of the initial 30cm3 of M plus some additional volume from the E.

To find the percentage of each gas, we can use the formula:

Volume of gas X / Total remaining volume * 100

Using the formula for M:
Percentage of M = (30cm3 / 61.5cm3) * 100 = 48.78%

Using the formula for E:
Percentage of E = (Remaining volume - Volume of M) / Total remaining volume * 100 = (31.5cm3 / 61.5cm3) * 100 = 51.22%

So, the percentage by volume of methane (M) is approximately 48.78% and the percentage by volume of ethane (E) is approximately 51.22%. Phew, that wasn't too explosive, was it?

2) Oh boy, another explosion to figure out the composition of a gas mixture. Let me grab my clown goggles and get to work!

We start with 32cm3 of a mixture of CO2, CH4, and H2. Let's represent them with their respective letters, so we have C, H, and O. We also have 50cm3 of O2. And after the explosion and cooling, we observe a decrease of 22cm3 in the total volume.

Hmm, sounds like we have an excess of O2. So, we can assume that the reaction consumed all the available fuel (C, H) and left some unreacted O2 behind.

To calculate the composition of the mixture, let's first find out how much O2 was used in the reaction:
Total volume decrease - Excess O2 volume = O2 used
22cm3 - 16cm3 = 6cm3

Now, let's figure out the volumes of CO2, CH4, and H2 in the mixture. Since there was an excess of O2, we know that it reacted with both carbon and hydrogen in CH4 and CO2.

Volume of CO2 = Volume of O2 used
Volume of CH4 = Volume of O2 used
Volume of H2 = 0 (assuming it completely reacted)

So, the composition of the mixture would be:
CO2: 6cm3
CH4: 6cm3
H2: 0cm3

That's it! Remember, don't try this at home unless you have clown supervision.

3) Alright, another explosion? I hope you're not doing these problems in your kitchen! Let's put on our clown safety gear and calculate the composition of this explosive mixture.

We start with 24cm3 of a mixture of CH4 (let's call it "M") and C2H6 (let's call it "E"). And we mix it with 90cm3 of O2. After the explosion and cooling, we observe a decrease of 32cm3 in the total volume when treated with KOH solution.

Let's find out how much O2 was used in the reaction:
Total volume decrease - 32cm3 = O2 used
32cm3 - 32cm3 = 0 cm3

Oh boy! It looks like there was no O2 used in this reaction. That means all the fuel (M and E) reacted completely.

To calculate the composition of the mixture, let's assume the reaction consumed all the available fuel.

Volume of M = 24cm3
Volume of E = 24cm3

So, the composition of the mixture would be:
CH4: 24cm3
C2H6: 24cm3

And there you have it! No guessing involved, just pure clown chemistry!

4) Alright, time to solve another explosive puzzle! Let's put on our clown lab coats and calculate the composition of this gas mixture.

We start with 195cm3 of a mixture of hydrogen (H2), carbon monoxide (CO), and nitrogen (N2). And we mix it with 150cm3 of oxygen (O2). After the explosion and cooling, the total volume decreases to 165cm3.

But wait, there's more! We also have another volume after absorption with KOH, which is 75cm3. So, it seems like we need to break down the volume changes step by step.

Step 1: Explosion volume change
Initial volume - Final volume = Volume change
195cm3 + 150cm3 - 165cm3 = 180cm3

Now, let's find out how much O2 reacted:
Total volume change - KOH volume = O2 used
180cm3 - 75cm3 = 105cm3

Step 2: Calculate the remaining volumes of H2, CO, and N2 after the reaction:
Volume of H2 = Volume of O2 used
Volume of CO = Volume of O2 used
Volume of N2 = Total volume - Sum of H2, CO, and O2 used

Volume of H2 = 105cm3
Volume of CO = 105cm3
Volume of N2 = 165cm3 - 105cm3 - 105cm3 = -45cm3 (Uh-oh, clown physics alert!)

Hmm, it seems we have a negative volume of N2. That doesn't seem right! Perhaps there was an error in the data or calculations, or we simply have a clown-caused explosion playfully bending the laws of physics.

In any case, it's time to call in the clown scientists for further investigation. Sorry, but it looks like our calculations aren't quite exploding with accuracy.

1) To find the percentage by volume of each gas in the mixture, we need to calculate the volume of each gas present before the explosion.

Let's assume x cm3 is the volume of methane (CH4) in the mixture.
Therefore, the volume of ethane (C2H6) in the mixture will be (30 - x) cm3.

Since the total volume of the mixture before the explosion is 30 cm3, we can write the total volume equation: x + (30 - x) = 30

Simplifying the equation, we have:
30 - x + x = 30
30 = 30

This means that the equation is valid, and we can proceed to calculate the percentage by volume of each gas present in the mixture.

The volume of methane (CH4) remaining after the explosion and cooling is 61.5 cm3, while the total volume before the explosion was 30 cm3.

The percentage of methane can be calculated as:
(61.5 cm3 / 30 cm3) × 100% ≈ 205%

Since the calculated percentage is greater than 100%, methane cannot be more than the initial volume (30 cm3). Therefore, there must be an error in the given data or calculations.

2) Similarly, to calculate the composition of the mixture, we need to determine the volume of each gas present before the explosion.

Let's assume x cm3 is the volume of CO2 in the mixture, y cm3 is the volume of CH4, and z cm3 is the volume of H2.

According to the given data, we know that:
Total volume = CO2 volume + CH4 volume + H2 volume
32 cm3 + 50 cm3 (O2) - 22 cm3 (reduced volume) = x cm3 + y cm3 + z cm3

We also know that there is an excess of 16 cm3 of O2, so:
16 cm3 = 50 cm3 (O2) - (reduced volume)
reduced volume = 34 cm3

Substituting these values into the equation, we get:
32 cm3 + 34 cm3 = x cm3 + y cm3 + z cm3
66 cm3 = x cm3 + y cm3 + z cm3

However, without additional information or constraints, it isn't possible to determine the composition of the mixture.

3) Without knowing their respective volumes in the mixture, it is not possible to calculate the composition of the mixture based on the given information.

4) To determine the percentage by volume of each gas present, we need to calculate the volume of each gas before the explosion.

Let's assume x cm3 is the volume of hydrogen (H2), y cm3 is the volume of carbon monoxide (CO), and z cm3 is the volume of nitrogen (N2) in the mixture.

According to the given data, the total volume equation can be written as:
x cm3 + y cm3 + z cm3 + 150 cm3 (O2) = 165 cm3 (final volume after cooling)

We also know that after absorption with KOH, the volume reduces to 75 cm3, so:
75 cm3 = x cm3 + y cm3 + z cm3

Now we have two equations:
x cm3 + y cm3 + z cm3 + 150 cm3 = 165 cm3
x cm3 + y cm3 + z cm3 = 75 cm3

By solving these equations simultaneously, we can find the values of x, y, and z, representing the volumes of each gas.

After finding x, y, and z, the percentage by volume of each gas can be calculated using the formula:
Percentage = (Volume of Gas / Total Volume) × 100%

To solve these questions, we need to use the concept of stoichiometry and the ideal gas law. The steps to find the composition of each gas in the mixture are as follows:

Step 1: Determine the moles of oxygen (O2) involved in the reaction.
Using the ideal gas law, we can determine the number of moles of O2 present in each case. The ideal gas law equation is:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

At room temperature and pressure (RTP), the pressure (P) is constant. The volume (V) and temperature (T) are given or can be assumed to be constant.

Step 2: Determine the moles of each gas involved in the reaction.
Let's assume that x moles of methane (CH4) are present and y moles of ethane (C2H6) are present.

From the balanced chemical equation, we know that:
2 moles of methane react with 7 moles of oxygen to form carbon dioxide and water.
2CH4 + 7O2 → 2CO2 + 4H2O

Similarly, 2 moles of ethane react with 7 moles of oxygen to form carbon dioxide and water.
2C2H6 + 7O2 → 4CO2 + 6H2O

Step 3: Using the stoichiometry, calculate the volume of carbon dioxide (CO2) formed in the reaction.
By using the stoichiometric ratios from the balanced equations, we can calculate the moles of CO2 produced for a given amount of oxygen reacted.

Since we have excess oxygen in all the cases, we will determine the volume of carbon dioxide formed by using the volume ratio.

Step 4: Using the volume of carbon dioxide formed, calculate the volume of remaining gases.
By using the stoichiometry from the balanced equation, we can determine the volume of each gas using the volumes of carbon dioxide formed and remaining.

For example, in the first question, if V1 is the volume of carbon dioxide produced, and V2 and V3 are the volumes of methane and ethane remaining, respectively, then we have the following relationship:
V1 + V2 + V3 = V_total - V_remaining

Using these steps and additional information given in each question, you should be able to determine the percentage by volume of each gas in the mixtures.