00:01
So this problem gives us the following set of equations.
00:04
Cf3 -cl plus uv light forms cf3 plus cl and then cl plus ozone forms c -l -2.
00:28
Ozone plus uv forms o2 plus ozone plus uv forms o2 plus o and then c l and o form c l plus o and now that we have c l we can cycle back to this equation and repeat that for every ozone molecule that we have.
01:00
And so based on this set of equations, we're given that we have 15 .0 grams of cf3 cl.
01:12
And we want to find the volume of ozone destroyed going through 10 cycles of the reaction that was on the previous page.
01:21
And so the first thing we want to do is convert grams of cf3 c .l into moles.
01:28
And so to do that, we use the molar mass of cf3cl, which is 104 .46 grams per mole.
01:39
And these are moles of cf3 cl, but we need moles of just cl.
01:46
And that is a one -to -one ratio.
01:49
So one mole of cf3 cl is one mole of cl.
01:55
And so we get our answer to be 0 .1436 moles of cl.
02:06
Now that we have moles of cl, we can find moles of ozone and then use the ideal gas law to find the volume of that ozone.
02:16
So we take our 0 .1436 moles of chlorine cl.
02:24
And the first thing we do is a mole -to -mole ratio...