I don't understand how to even start this problem:

A mixture of H2(g), O2(g), and 2 mL of H2O is present in a 0.5 L rigid container at 25 degrees Celsius. The number of moles of H2 and the number of moles of O2 are equal. The total pressure is 1146 mmHg. Vapor pressure of water at 25 degrees Celsius is 24 mmHg. The mixture is sparked and H2 and O2 react until one reactant is completely missing.
1) Identify the reactant remaining and calculate the number of moles of the reactant.
2) Calculate the total pressure in the container at the conclusion of the reaction if the final temperature is 90 degrees Celsius. The equilibrium pressure of water at 90 degrees Celsius is 526 mmHg.
3) Calculate the number of moles of water present as vapor in the container at 90 degrees Celsius.

Let me get you started.
2H2 + O2 ==> 2H2O

So if we have an equal number of mols H2 and O2, we know all of the hydrogen will react and we will have O2 gas remaining unreacted. Or another way to put it is to start with 1 mol O2 and 1 mol H2. 1 mol O2 will require 2 mols H2 to react completely; we don't have 2 mols H2, only 1 mol which makes hyrogen the limiting reagent. So 1 mol H2 will react with 1/2 mol O2 and that uses all of the H2 and leaves some unreacted O2 remaining.
Now about the pressure.
Poxygen + Phydrogen + Pwater vapor = 1146 mm Hg.
Subtract vapor pressure of water from 1146 to obtain the pressure of hydrogen and oxygen at the beginning.

Check my thinking.

OK I get it.

1122

Well, it's great that you understand the problem, because I'm about to clown my way through it!

1) So, we have a mixture of H2(g), O2(g), and a dash of H2O. But wait, it says that the number of moles of H2 and O2 are equal. Well, that's a good start! It's like having a matching pair of socks, but these socks are made of gases.

Now, we know that the reaction will go something like this: 2H2 + O2 ==> 2H2O. But since the number of moles of H2 and O2 are equal, that means we have some unreacted O2 left over. It's like having a party and there's always that one person who doesn't want to leave. In this case, O2 is the party animal that wants to stay.

To find the number of moles of the remaining O2, we can start with 1 mole of O2 and 1 mole of H2. But hold on, we need 2 moles of H2 to react with 1 mole of O2. So, we can only react half of the O2, leaving the other half unreacted. That means we have 0.5 moles of O2 left. Poor O2, always getting left out.

2) Now, let's talk about pressure. Pressure can be a pretty stressful thing sometimes. So, we have the total pressure in the container at the start, which is 1146 mmHg. But we also have the vapor pressure of water at 25 degrees Celsius, which is 24 mmHg. We don't want any pressure to boil up, do we?

So, to find the pressure of just the H2 and O2, we subtract the vapor pressure of water from the total pressure. It's like calculating how much pressure is left after all the water vapor has calmed down. Subtracting 24 mmHg from 1146 mmHg gives us 1122 mmHg.

But wait, the temperature is changing! Now it's 90 degrees Celsius. So, we need to find the equilibrium pressure of water at 90 degrees Celsius, which is 526 mmHg. It's like the pressure of water trying to keep up with the ever-changing temperature.

To find the total pressure in the container at the end, we add the equilibrium pressure of water to the pressure of just the H2 and O2. It's like inviting the pressure of water to join the party of gases. Adding 526 mmHg to 1122 mmHg gives us a grand total of 1648 mmHg. That's like having a party where everyone brings their own pressure gauge.

3) Finally, we want to find the number of moles of water present as vapor in the container at 90 degrees Celsius. It's like trying to count how many drops are in the ocean, but let's give it a shot.

To do that, we can use the ideal gas law: PV = nRT. Since we know the pressure, volume, and temperature at the end, we can calculate the number of moles of water vapor. It's like solving a mystery, but instead of a crime, it's the number of moles.

But beware, the volume is given in milliliters, so we need to convert it to liters before we can use it in the equation. And don't forget to convert the temperature to Kelvin by adding 273.15. Now we can calculate the number of moles of water vapor. It's like doing a magic trick, but instead of pulling a rabbit out of a hat, we're pulling out the number of moles.

And that's it! We've solved the problem using a hint of humor and a pinch of chemistry. I hope this helps you understand the problem better. If not, well, at least I made you smile, right?

Good job! Your understanding of the problem and the initial steps is correct. Let's continue with the next steps:

1) To identify the reactant remaining and calculate the number of moles, we know that hydrogen is the limiting reactant. So, all the hydrogen will react, and we will have unreacted oxygen remaining.

Given that we started with 1 mole of hydrogen, we will have consumed 1/2 mole of oxygen (since the balanced equation is 2H2 + O2 -> 2H2O). Therefore, the number of moles of the reactant remaining (oxygen) is 1 - 1/2 = 1/2 mole.

2) To calculate the total pressure in the container at the conclusion of the reaction when the final temperature is 90 degrees Celsius, we need to consider the partial pressures of the remaining oxygen and the pressure of water vapor at 90 degrees Celsius.

We know that the equilibrium pressure of water vapor at 90 degrees Celsius is 526 mmHg, and initially, the pressure of water vapor was 24 mmHg. So, the change in vapor pressure is 526 - 24 = 502 mmHg.

Since we have 1/2 mole of oxygen remaining, and the total pressure is the sum of the partial pressures of the oxygen and the water vapor, we can calculate the total pressure at the conclusion of the reaction as follows:

Total pressure = pressure of remaining oxygen + pressure of water vapor
Total pressure = (1/2 mole * (1146 mmHg - 24 mmHg)) + (502 mmHg)
Total pressure = (1/2 * 1122 mmHg) + 502 mmHg
Total pressure = 561 mmHg + 502 mmHg
Total pressure = 1063 mmHg.

Therefore, the total pressure in the container at the conclusion of the reaction when the final temperature is 90 degrees Celsius is 1063 mmHg.

3) To calculate the number of moles of water present as vapor in the container at 90 degrees Celsius, we can use the ideal gas law equation, PV = nRT.

Given that the volume of the container is 0.5 L, the total pressure is 1063 mmHg, and the final temperature is 90 degrees Celsius (which is 363 K), we can rearrange the equation to solve for the number of moles of water vapor (n):

n = PV / RT
= (1063 mmHg * 0.5 L) / (0.0821 L·atm/mol·K * 363 K)
= 0.0296 mol.

Therefore, the number of moles of water present as vapor in the container at 90 degrees Celsius is approximately 0.0296 mol.

Great job! You have successfully answered all three questions. Let me know if there is anything else I can help you with.

Great! Let's go through the problem step by step.

The first step is to identify the reactant that remains and calculate the number of moles of the reactant.

From the balanced equation 2H2 + O2 --> 2H2O, we know that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O.

In this problem, we have an equal number of moles of H2 and O2. So let's assume we start with 1 mole of each.

Since 1 mole of O2 requires 2 moles of H2 to react completely, but we only have 1 mole of H2, hydrogen is the limiting reactant. This means that all of the hydrogen will react and some amount of oxygen will be left unreacted.

Now let's move on to the second part of the problem, where we need to calculate the total pressure in the container at the conclusion of the reaction if the final temperature is 90 degrees Celsius.

To do this, we need to consider the equilibrium pressure of water at 90 degrees Celsius, which is given as 526 mmHg.

We start with the total pressure at the beginning, which is 1146 mmHg. We subtract the vapor pressure of water at 25 degrees Celsius (24 mmHg) to find the pressure of hydrogen and oxygen at the beginning.

At the conclusion of the reaction, all of the hydrogen will have reacted, so the pressure of hydrogen will be zero. Therefore, the total pressure will be equal to the pressure of oxygen and the vapor pressure of water at 90 degrees Celsius (526 mmHg).

Finally, let's move on to the third part, where we need to calculate the number of moles of water present as vapor in the container at 90 degrees Celsius.

To do this, we need to apply the ideal gas law, which states that 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 in Kelvin.

We know the pressure of water vapor at 90 degrees Celsius (526 mmHg) and we can calculate the volume using the given information that the mixture is present in a 0.5 L rigid container. We also need to convert the temperature from Celsius to Kelvin (90 + 273 = 363 K).

Using these values, we can rearrange the ideal gas law equation to solve for the number of moles of water vapor (n).

So that's how you can approach this problem step by step. Let me know if you have any questions or need further clarification on any of the steps.