00:01
Alright, so here we are given our complete combustion reaction for our ethane.
00:05
Our theoretical amount of oxygen that is required is 3 .5 moles of oxygen per mole of ethane.
00:12
Given that our flow rate is 100, our theoretical amount, we can take our 3 .5 times 100 to get 350.
00:22
However, our problem states that there is 20 % excess of oxygen.
00:27
Therefore, our actual amount of oxygen is going to be 1 .2 times 350, which is going to give us 420.
00:37
Now we can use the ideal gas law to calculate our volumetric flow rate of oxygen.
00:44
This is given by pv is equal to nrt.
00:50
Rearranging this for v, we're going to get that v is equal to nrt divided by p.
01:01
We can substitute in our given values.
01:05
We're going to get 420 times 0 .0821 times 175 plus 273 divided by 1 .1.
01:19
Therefore, we're going to get our volumetric flow rate of our oxygen is 140...