00:02
To find the limiting reactant, first we need to divide each of the reactant masses by their respective molar masses.
00:09
In the case of this platinum compound, that's 415 .09 grams in every one mole, which is equal to 0 .134 moles of this compound.
00:26
We're going to do the same for the ammonia, find how many moles is that we have.
00:30
That's equal to, its molar mass is equal to 17 .03 grams in every one mole.
00:41
So we find that this is equal to 2 .09 moles.
00:46
Remember that these exist in 1 to 2 mole ratios.
00:52
So for every one mole of the platinum compound, i need 2 moles, twice as many moles, of ammonia.
00:59
Clearly, two moles is much, much bigger than twice this number of moles.
01:07
We can do the math that this many moles of the platinum compound needs 0 .268 moles of ammonia.
01:19
We have way more ammonia than that, and that's how we determine that the platinum compound is the limiting reaction.
01:28
Reactant, and thus we use that to determine how much product is produced...