00:01
Hi, so to solve for this, we're already given with the balance equation and we're given with both amounts of the reactants.
00:09
That means we have to determine first the limiting reactant.
00:13
Let's convert the given mass of reactants to moles.
00:16
We'll start with ammonia.
00:17
We have 25 grams of ammonia.
00:19
Convert to moles by dividing the molar mass of this compound.
00:22
That's 17 .04 grams.
00:24
And we have here moles.
00:26
So that means we could cancel grams here and this will give us 1 .467.
00:34
This is moles of nh3.
00:38
And then for oxygen gas, we have 45 grams.
00:43
Convert to moles by dividing the molar mass of o2.
00:47
That's 32 grams.
00:48
And then we have here moles.
00:50
So we could cancel grams.
00:52
45 divided by 32 is 1 .406 moles of o2.
00:59
And then we will solve for mole ratio.
01:03
Mole ratio will be equivalent to the number of moles of the reactant divided by the coefficient of that reactant from the balance equation...