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

Question

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?

Answer 1

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

Answer 2

duh, that makes sense. Thanks

Answer 3

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.

Answer 4

sweg

Answer 5

12l

Answer 6

The answer is 12 L.