00:02
Hi there! in this question we are given the mass of our reactant and we are trying to determine the mass of one of our products that is formed, making this a mass -to -mass stoichiometry problem.
00:23
And the first thing we need in any type of stoichiometry problem is a balanced chemical equation.
00:28
So let's jump in with our unbalanced equation here and start balancing it.
00:33
All right, so we see we have one potassium on the left but two on the right.
00:37
So we'll put a 2 in front of kno3.
00:40
That gives us 2 nitrogen and we already have 2 nitrogen, that's awesome.
00:44
That gives us 2 times 3 or 6 oxygen.
00:48
Well on the right side we have 3, which is an odd number.
00:53
So we could put numbers in front of o2 but we're still going to have to add that other oxygen giving us an odd number.
01:00
So we know there is no way to make these equal unless i go ahead and make an even number of oxygen on the right side.
01:09
So i'm going to put a 2 in front of k2o because now i have an even number of oxygen to work with.
01:15
But of course that's going to change our previous balancing.
01:19
So going back to the left side now i need a 4 because i need 4 potassium since i have 4 potassium on the right.
01:26
I have 4 nitrogen so i now need a 2 in front of n2 to give me 4 nitrogen on the right.
01:33
On the left now i have 4 times 3 or 12 oxygen.
01:36
There are 2 oxygen in the 2k2o.
01:40
That means i need another 10 from the o2.
01:43
So i need to put a 5 in front of the o2.
01:46
This is now balanced with the smallest number of whole number coefficients.
01:51
We can't reduce these and still have whole numbers.
01:54
So this is our balanced equation and we are given 58 .6 grams of the kno3.
02:02
And we want to know how many grams of the nitrogen gas are formed.
02:07
As i mentioned this is a mass to mass stoichiometry problem.
02:10
And those will follow 3 prescribed steps.
02:14
The first is to convert the given into moles of the given.
02:18
And we do that by using the molar mass.
02:20
For the kno3 it's given to us in the problem.
02:25
Then we can convert from moles of the kno3 to moles of the nitrogen by using the mole ratio.
02:32
The mole ratio comes from the coefficients in the balanced equation.
02:35
So we see that the mole ratio here is a 4 to 2...