Create an engaging image that encapsulates the process of a chemistry problem. The scene features various laboratory apparatus such as test tubes, a Bunsen burner, and professional measuring equipment. Chemical symbols like O2, CS2, CO2, and S2 float in the air, hinting at a chemical equation. Nearby, a mole, a unit of measure in chemistry, sneaks into the picture, signifying the calculation of mols in the equation. Lastly, frames with five different volumes of gas containers labeled from 'A' to 'E', each representing the potential outcomes of the problem, appear at the foot of the scene.

What volume of O2 is required to react with excess CS2 to produce 4.0L of CO2 Assume all gases at STP.

Not sure what I am doing wrong, but the answer I am getting is not one of the answers.

4L/X mole=22.4 L/1 mole ===>.1786 mols
Balanced equation:
O2 + CS2 --> CO2 + S2
.1786 mols of each since it is a 1-1-1-1 equation.

Therefore, I am getting .1786 mols of O2 to make .1786 (4 L) CO2. .1786 mols O2 is 4 liters.

My answer options are
A) 12 L
B) 22.4 L
C) 1/3 x 22.4 L
D) 2 x 22.4 L
E) 3 x 22.4 L

What am I doing wrong here?

Just a note to clarify a couple of points.

#1. IF S were the product, and probably it is not, you would write it as S8. SO2 is the more likely product.

#2. I notice you went through mols. When all of the products are gases, you can use a short cut and not go through mols. The usual way of doing this is to say
Step 1.
4.0 L CO2 x (1 mol/22.4 L) = 0.1786 mols CO2.

Step 2.
mols O2 = 0.1786 mols CO2 x (3 mols O2/1 mol CO2) = 0.5357 mols O2.

Step 3.
volume O2 = 0.5357 x (22.4 L/1 mol) = 12 L

But here is the shortcut. Notice that we divided by 22.4 in step 1 and multiplied by 22.4 L in step 3. So all we need is step 2.
4.0 L CO2 x (3 mols O2/1 mol CO2) = 12 L.

sweg

I have a hunch that what you wrote as S2 would be 2S if pure sulfur were indeed the result of such a reaction but I have a further hunch that SO2 or something would be a product.

For example (and my chemistry is terrible and I loaned someone my book):
maybe
3O2 + CS2 --> CO2 + 2SO2
in that case I would need 3*4 or 12 L

12l

The answer is 12 L.

To determine the correct volume of O2 required to react with excess CS2, let's break down the problem step by step:

1. Start with the given information:
- Volume of CO2 produced: 4.0 L
- All gases are at STP (standard temperature and pressure), which means 1 mole of any gas occupies 22.4 L.

2. Find the moles of CO2 produced:
Using the given volume of CO2 (4.0 L) and the relationship between volume and moles at STP (22.4 L = 1 mole), we can calculate the moles of CO2 produced:
Moles of CO2 = Volume of CO2 / 22.4 L
= 4.0 L / 22.4 L
= 0.179 mol (rounded to three decimal places)

3. Determine the stoichiometry of the balanced equation:
The balanced equation is: O2 + CS2 --> CO2 + S2
From the equation, we can see that the mole ratio between O2 and CO2 is 1:1, meaning that 1 mole of O2 reacts to produce 1 mole of CO2.

4. Calculate the moles of O2 required:
Since the mole ratio between O2 and CO2 is 1:1, the moles of O2 required will be the same as the moles of CO2 produced:
Moles of O2 = Moles of CO2
= 0.179 mol

5. Convert moles of O2 to volume of O2 at STP:
Using the mole-volume relationship at STP (22.4 L = 1 mole), we can calculate the volume of O2 required:
Volume of O2 = Moles of O2 × 22.4 L/mol
= 0.179 mol × 22.4 L/mol
≈ 4.0 L (rounded to one decimal place)

The correct answer is 4.0 L of O2. It seems your calculations were correct, and you arrived at the same answer as well. However, none of the provided answer options match the correct answer. In this case, it appears there may have been an error in the given answer choices.

duh, that makes sense. Thanks