posted by T on .
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
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?
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):
3O2 + CS2 --> CO2 + 2SO2
in that case I would need 3*4 or 12 L
duh, that makes sense. Thanks
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
4.0 L CO2 x (1 mol/22.4 L) = 0.1786 mols CO2.
mols O2 = 0.1786 mols CO2 x (3 mols O2/1 mol CO2) = 0.5357 mols O2.
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.