00:01
Hi, so in this case we're going to have 54 grams of aluminum and 24 .5 grams of o2.
00:12
Each of these correspond to, in amount of mole of aluminum, we need to divide the mass of aluminum by the molar mass of the compound, right? so the molar mass of aluminum is equal to 26 .98, let's round it up to 27 grams per mole.
00:35
This means that we have approximately 2 moles of aluminum and in the case of o2 this is going to be 24 .5 grams divided by 32 grams per mole.
00:52
So this gives us 0 .766, 766 moles of o2.
01:03
Next up let's look at our reaction equation.
01:08
We have three equivalents of oxygen gas plus two equivalents of aluminum oxide, sorry, plus four equivalents of aluminum alone as a solid, are going to react and produce two equivalents of aluminum 2o3, the aluminum oxide as a solid.
01:36
And we're given the delta h of the reaction which is minus 3339 .6 kj per mole of reaction.
01:49
Okay, so now the next thing that we need to do is first of all determine which one is going to be the limiting reagent.
01:56
Here we can see that if we take oxygen as the limiting reagent this amount of moles correspond to 0 .766 mole of o2...