1. If 29.0 L of methane, CH4, undergoes complete

combustion at 0.961 atm and 140°C, how many liters
of each product would be present at the same
temperature and pressure?

My answer:
(0.961 atm)(29.0 L) (413k) all divided by (413)(0.961 atm)=29.0 L CO2

29.0 L CO2 x 2LH2O/1LCO2=58.0 L H2O

2. If air is 20.9% oxygen by volume,
a. how many liters of air are needed for complete
combustion of 25.0 L of octane vapor, C8H18?
b. what volume of each product is produced?

****I do not know how to do a, but I got 200L CO2 and 225 L H2O vapor for part b.

3.Methanol, CH3OH, is made by causing carbon
monoxide and hydrogen gases to react at high
temperature and pressure. If 4.50 × 102 mL CO and
8.25 × 102 mL H2 are mixed,
a. which reactant is present in excess?
b. how much of that reactant remains after the
reaction?
c. what volume of CH3OH is produced, assuming the
same pressure?

*****a) the reactant in excess is CO because it produces 14 g CH3OH and H2 produces 13.2 g CH3OH...14>13.2

I need help on b) and c)...

I might be asking you some questions after you helped me, so keep track of my post....in other words,there might be some follow-up questions.....

1. If 29.0 L of methane, CH4, undergoes complete

combustion at 0.961 atm and 140°C, how many liters
of each product would be present at the same
temperature and pressure?

My answer:
(0.961 atm)(29.0 L) (413k) all divided by (413)(0.961 atm)=29.0 L CO2
I really don't understand what you've done here but 29.0L is the correct answer. The easy way to work this problem is to note that if everything is in the gaseous state then you need not convert everything to mols, do the mols conversion, then back to L. Instead you can use L as mols; therefore, 29.0L CH4 produces 1 mol CO2 for each mol CH4 and that is 29.0L CO2.
For H2O you have done it right; i.e., 1 mol CH4 produces 2 mols H2O; therefore, 29.0L x 2 = 58.0 L H2O (in the gaseous state of course).


29.0 L CO2 x 2LH2O/1LCO2=58.0 L H2O

2. If air is 20.9% oxygen by volume,

a. how many liters of air are needed for complete
combustion of 25.0 L of octane vapor, C8H18?
b. what volume of each product is produced?

****I do not know how to do a, but I got 200L CO2 and 225 L H2O vapor for part b.

You do this the same way, up to a point, that you solve for CO2 and H2O (the 200L and 225L values are correct).
25L C8H18 x (25 mol O2/2 mol C8H18) = 312.5 L O2 (if it were pure O2).
Then 20.9% of what number = 312.5
0.209x = 312.5
x = about 1495L at 20.9% purity. Watch the number of significant figures; you're not allowed 4 places.

3.Methanol, CH3OH, is made by causing carbon

monoxide and hydrogen gases to react at high
temperature and pressure. If 4.50 × 102 mL CO and
8.25 × 102 mL H2 are mixed,
a. which reactant is present in excess?
b. how much of that reactant remains after the
reaction?
c. what volume of CH3OH is produced, assuming the
same pressure?

*****a) the reactant in excess is CO because it produces 14 g CH3OH and H2 produces 13.2 g CH3OH...14>13.2

I need help on b) and c)...

I'm assuming high T and P means everything is gaseous. You're right with a that H is the limiting reagent (LR) and CO is the excess reagent (ER). You converted to grams but you can keep everything in the gaseous state and stick with mL.

b.
If H2 is the LR, then how much CO will be used for complete reaction? That is
825 mL H2 x (1 mol CO/2 mol H2) = 412.5 mL CO used.
We had 450 initially so 450-412.5 = ? mL left unreacted.

c.
Assuming high T and P (so CH3OH is a gas) the reaction will produce 825 mL H2 x (1 mol CH3OH/2 mol H2) = 825 x 1/2 = ? mL CH3OH VAPOR.
Again, watch the number of s.f. You're allowed 3. from 8.25E2 and 4.50E2.

If you have follow up questions please

a. post to my attention AND THIS IS IMPORTANT
b. Copy the problem and answers followed by your follow up question; otherwise, this post can get lost in the shuffle and I'll not be able to find it.

No problem, I'm here to help with your questions! Let's tackle them one by one.

For question 1, it looks like you've already determined that 29.0 L of CH4 will produce 29.0 L of CO2 and 58.0 L of H2O.

For question 2a, to find out how many liters of air are needed for complete combustion of 25.0 L of octane vapor, we can use the fact that air is 20.9% oxygen by volume. So, we need to calculate how much air contains 20.9% oxygen by volume and is equal to 25.0 L of octane vapor.

Let's set up the equation:
25.0 L octane vapor x (1 L air / 20.9 L oxygen) = Answer

For part b, where you got 200 L CO2 and 225 L H2O vapor, it seems like you might have just calculated the volumes of the products instead of the individual volumes of CO2 and H2O. Since complete combustion of octane produces 8 molecules of CO2 and 9 molecules of H2O, you'll need to multiply each volume by the respective ratios.

For question 3:

a) To determine which reactant is in excess, you can compare the given volumes of CO and H2. Since 1 mole of CO reacts with 1 mole of H2 to produce 1 mole of CH3OH, we'll need to convert the given volumes to moles and see which reactant has excess moles.

b) To find out how much of the excess reactant remains after the reaction, you'll need to calculate the moles of the limiting reactant (the one that is completely consumed) by using stoichiometry.

c) To calculate the volume of CH3OH produced, you'll need to convert the moles of CH3OH obtained in part b) to volume using the ideal gas law, assuming the same pressure.

Please let me know which specific part you need help with, and I'll be happy to assist you!

Sure, I'm here to help with your questions. Let's address them one by one.

1. For the combustion of methane, CH4:
To find the number of liters of each product, we need to use the balanced chemical equation for the combustion of methane, which is:
CH4 + 2O2 -> CO2 + 2H2O

According to the stoichiometry of the reaction, 1 mole of methane produces 1 mole of carbon dioxide (CO2) and 2 moles of water (H2O). The coefficients of the balanced equation tell us the ratio of moles, which can then be converted to liters using the ideal gas law.

The given volume of methane is 29.0 L. We can convert this to moles using the ideal gas law equation:
moles = volume / molar volume
where the molar volume at standard temperature and pressure (STP) is 22.4 L/mol.

moles of CH4 = 29.0 L / 22.4 L/mol ≈ 1.295 moles

From the balanced equation, we can see that 1 mole of methane produces 1 mole of carbon dioxide. Therefore, the number of moles of carbon dioxide produced is also approximately 1.295 moles.

Now, using the ideal gas law equation again:
volume = moles * molar volume

volume of CO2 = 1.295 moles * 22.4 L/mol ≈ 29.0 L

So, the volume of carbon dioxide produced is approximately 29.0 L.

Next, we can use the stoichiometry of the balanced equation to find the volume of water produced. We know that for every 1 mole of methane, 2 moles of water are produced.

moles of H2O = 1.295 moles * 2 = 2.59 moles

Now, using the ideal gas law equation once more:
volume of H2O = 2.59 moles * 22.4 L/mol ≈ 58.0 L

So, the volume of water produced is approximately 58.0 L.

Therefore, at the same temperature and pressure, there would be 29.0 L of carbon dioxide and 58.0 L of water present.

2. For the combustion of octane, C8H18:
a. To find the volume of air needed for complete combustion, we need to consider the balanced chemical equation for the combustion of octane, which is:
C8H18 + 12.5O2 -> 8CO2 + 9H2O

According to the stoichiometry of the reaction, 1 mole of octane requires 12.5 moles of oxygen (O2) for complete combustion. Given that air is 20.9% oxygen by volume, we can calculate the volume of air needed as follows:

moles of octane = volume of octane / molar volume of octane
moles of oxygen = moles of octane * 12.5
volume of air = moles of oxygen / mole fraction of oxygen in air

To calculate the mole fraction of oxygen (O2) in air, we convert the percent composition to decimal format:
mole fraction of oxygen = 20.9% / 100% = 0.209

Therefore, the volume of air needed can be calculated as:
volume of air = (volume of octane / molar volume of octane) * (moles of octane * 12.5) / 0.209

b. To find the volume of each product, we can use the stoichiometry of the balanced equation as before and calculate the moles of carbon dioxide (CO2) and water (H2O) produced. Then, convert the moles to volume using the ideal gas law.

3. For the reaction between carbon monoxide (CO) and hydrogen (H2) to produce methanol (CH3OH):
a. To determine which reactant is in excess, we need to compare the stoichiometry of the balanced equation to the given amounts of CO and H2. The balanced equation is:
CO + 2H2 -> CH3OH

We have 4.50 x 10^2 mL of CO and 8.25 x 10^2 mL of H2. We can convert these volumes to moles using the ideal gas law equation and molar volumes at STP.

b. To determine how much of the excess reactant remains after the reaction, we need to calculate the moles used in the reaction and subtract it from the initial moles of the excess reactant.

c. To find the volume of methanol produced, we can use the stoichiometry of the balanced equation. The number of moles of methanol produced can be calculated using the moles of the limiting reactant and the balanced equation coefficients. Then, using the ideal gas law equation, we can convert moles to volume.

I hope this clarifies the steps involved in solving these problems. Let me know if there's anything else I can help with or if you have any follow-up questions.