00:01
This question wants us to determine how many grams of oxygen are required for the complete combustion of 421 grams of ethanol.
00:11
So before we start this question, we must first calculate the molecular weight of ethanol and of oxygen gas.
00:24
So for oxygen gas, it's fairly straightforward.
00:28
It is just twice the weight of the oxygen atoms.
00:32
Which gives a total molecular weight of 32 grams per mole.
00:37
And the molecular weight for ethanol is twice the weight of the carbon atom, plus five times the weight of the hydrogen atom.
00:53
And the weight of the carbon atom should be 12 .01.
00:59
Actually, it should be six times the weight of the hydrogen atom, as hydrogen is present in ethanol six times.
01:06
Plus the weight of the oxygen atom, and that gives us a total molecular weight of 46 .08 grams per mole.
01:21
Now, the question tells us that the 421 grams of ethanol are being completely consumed...