Hey Im cant find the answer for this question, can someone tell me how to do it?
What volume of air (21.0% oxygen, O2, by volume), measured at SLC, would be required to fully combust 68.5 g of C3H8?
thank youuu
Hi!
So to figure out this question your first step would be to create a balanced equation of the complete combustion of C
3H
8.
The equation would look like this [I've emitted states to save time, but please don't forget states in questions].
C
3H
8 + 5O
2 -> 3CO
2 + 4H
2O
Now, we know that we have 68.5 grams of Propane. Therefore, to find out how much oxygen is needed to react with this amount of propane to make it combust, we need to find out how many mols of Propane we have.
Using n=m/fm
68.5/44 = 1.56mol
Next, we know we need 5 times as much O2 to react with this amount of propane, using the chemical equation. Hence,
n(O2) = 1.56 * 5 = 7.8 mol.
Then, as this question is asking for Volume, we need to find out how much Litres of O2 we need. As the experiment is being conducted under SLC, we use n = V/Vm.
V = 7.8 *24.8 = 193.44 L
Now, finally, we need to find out how much air gives us this much of oxygen. 21% of air is oxygen, so we can form an equation like this and rearrange.
V(air) * 0.21 = 193.44
V = 921.14 L of air.
I hope this helps! Let me know if you have more questions! I've attached my rough working below in case that helps better.