00:01
So for this problem, we have the decomposition reaction of limestone, and we want to find how many grams of quicklime can be produced.
00:09
So the first thing we want to do is we want to write the decomposition reaction of limestone.
00:13
So we know we have limestone as our only reactant, and we know that we have quick lime, as well as carbon dioxide, as our two products.
00:23
And immediately, when we look at the equation, we see that everything is balanced.
00:26
One calcium on each side, one carbon on each side, and three oxygen on each side.
00:31
So now that we have the balanced equation, we want to find the molar masses of the relevant compounds.
00:37
And in this case, it's going to be limestone as well as quicklime.
00:42
So let's first do limestone.
00:44
C .a .c .03.
00:45
Well, we know for the calcium, that's 40 .08.
00:51
For the carbon, it's 12 .01.
00:53
And with three oxygen atoms, we multiply 16 .00 times 3.
00:57
And then we get at the final more mass is 100 .07 grams or 100 .09 grams per mole.
01:14
So that's the molar mass of limestone.
01:16
Then we want to do quick lime...