00:01
So we'll start by writing our balanced equation.
00:03
We've got two lithium plus bromine to produce two libr properly balanced.
00:13
We're told that we're going to start with 25 grams of lithium, 25 grams of bromine.
00:18
So let's work out our massive libr from both sources.
00:23
So starting with 25 grams of lithium, 6 .9 grams per mole.
00:31
Here's our stochialometry, 2 moles of lithium to 2 moles of l -i -b -r, and 1 -mol.
00:44
Lithium bromide has a molar mass of 86 .8 grams of l -i -b -r.
00:50
Working this out will give me a mass of 314 grams of lithium bromide.
00:56
Let's do a similar calculation starting from 25 grams of the second reactants, br2.
01:01
Molar mass is 159 .8 grams and one mole...