00:01
Hello, here we have to calculate the theoretical ratio air to fuel on a mass and mall basis for combustion of ethanol.
00:09
But first of all, let us calculate a weight percentage of oxygen in air.
00:21
So, air contains 79 mole percent of nitrogen and 21 mole.
00:37
Percent of oxygen so let's presume that we have one mole of oxygen that's why this it weights 32 grams of oxygen so therefore number of moles of nitrogen then equals to one mole divided by 0 .21 multiplied by 0 .17 so let's calculate it.
01:22
There is 3 .76 mole of nitrogen and that's why mass of nitrogen equals to 3 .76 mole times 28 grams per mole which is 105 grams.
01:46
So therefore weight percentage of oxygen equals to 32 grams divided by 32 grams times 105 grams 3 plus 105 grams times 100 % so that is 0 .20 sorry that is 23 .4 percent okay now let's write down the reaction for the combustion of ethanol ethanol vx with oxygen producing co2 and water let's balance this equation so let's see how we can balance it let's add 2 before co2 and so now carbon is balanced let's see how we could have balanced hydrogen so to actually to balance hydrogen yeah probably it it's better for us to add 2 before ethanol now we have to add 4 before co2 and 3 before water and maybe it's even better for us to add 8 before co2 4 before ethanol now we can add 6 before water and now we can balance oxygen.
04:01
Now let's look at the mall ratio.
04:04
Yeah, first let's do the mall ratio.
04:18
So here we have so if we have nine more, sorry, four more of, four more of co2 plus nine more of oxygen, then we would have have four more of co2 plus nine more of oxygen, then we would have have four more of co2 plus, 9 over 0 .21 mall of air which would have been so that is formal yeah of course oh here of course oh here of course yeah it's ethanol so let's label it ethanol for now and therefore yeah we have four more of c2 oh actually yeah i think i've made a typo in the ethanol formula let me yeah let me double check the balancing yeah probably yeah the balancing should be different here let me double check my typo okay, now let's try to balance in the following way so let's add 2 before co2 3 before water and see how it could have balanced oxygen so obviously that's one here now let's calculate hydrogen, sorry, let's calculate oxygen.
06:42
So here it looks like we need to have three before oxygen...