00:01
We have four stoichiometry problems we've been asked to perform.
00:03
In each case, doing a synthesis reaction, starting with 3 .67 grams of a limiting reactant, and then finding how much product can be produced in each case.
00:16
So for the first one, we have the reaction.
00:20
Barium reacting with chlorine gas yields barium chloride.
00:27
And we're told that we are going to begin with 3 .6.
00:30
Six, seven grams of barium.
00:34
And we want to find the product.
00:37
So, we need to go to a mole ratio, and we want to find mass in grams of the product.
00:42
So we'll have to pass through our mole ratio, and then get back to using molar mass.
00:47
Now, if you look at the periodic table, you'll see that barium has a molar mass of 137 .3 grams for every mole.
00:57
Now, if you look at all the coefficients here, you'll see it's one to one to one.
01:03
So our mole ratio is going to be quite simple.
01:05
It'll be a mole of barium for every mole of product.
01:13
And then we'll have to use our molar mass.
01:16
So for a mole of barium chloride, we'll have to add up.
01:19
Barium has a molar mass of 137 .3.
01:23
Chlorine has a mass of 35 .45, but we have two of them.
01:29
So we'll add all that up, and it will come out to be 208 .2 grams of barium chloride.
01:36
Now if this is done correctly, all of the units should cancel except for mass of barium chloride.
01:43
So let's check that it does, and it looks like it will.
01:48
So we'll take our 3 .67 times our 208 .2 and divide by 137 .3.
01:55
And once this is done, it'll come out to be 5 .57 grams of barium chloride.
02:01
And we know this has to have three significant figures since we were given three to begin with, 3 .67.
02:09
On part b, we have a new reaction.
02:14
It's calcium oxide, reacts with carbon dioxide to form the ternary compound calcium carbonate.
02:23
And once again, we're starting with our first reactant and assuming that the co2 is excess.
02:29
We'll begin again with 3 .67 grams of calcium oxide.
02:33
So we'll follow the exact same steps as before.
02:36
We'll transfer to moles, do a mole ratio, and then get back to mass.
02:41
Of our product.
02:43
Now the molar mass of calcium is 40 .1 and the molar mass of oxygen is 16 grams per mole so we'll add those together and get 56 .1 grams of calcium oxide for every mole of this substance.
03:01
Once again we have a one to one to one ratio so our mole ratio will be quite simple.
03:08
For every mole of calcium oxide there is a mole of calcium carbonate.
03:19
And then we need mass of our product, so again we'll add this up.
03:24
The calcium is 40 .1.
03:28
The carbon is 12 and the oxygen is 16, but there's three of them.
03:35
So adding all that up, we'll get it to be 100 .1 grams of calcium carbonate.
03:46
Let's check our units again.
03:48
Grams of calcium oxide cancel each other out.
03:51
Moles of calcium oxide cancel each other out.
03:53
And moles of calcium carbonate cancel each other out...