00:01
Hi, so we are solving for the volume of oxygen gas that is produced when we have 59 .6 grams of mercury oxide.
00:10
So first we need to write the balance equation for this reaction.
00:15
Mercury to oxide, that's hgo, and this will decompose to mercury and oxygen gas.
00:23
So to balance this equation, we need to write 2 here and then 2 in here.
00:29
And then the next step is to solve for the number of moles of oxygen gas produced.
00:38
If we have 59 .6, this is 59 .6 grams of mercury oxide.
00:51
Now we have to convert this to moles, and we'll do that by dividing the molar mass of hgo.
00:56
So that's 200 .59 plus 16, the atomic mass of oxygen.
01:00
And we'll get 216 .59.
01:04
So you just add the atomic mass of mercury from the predictable and the atomic mass of oxygen.
01:10
So this is moles of hgo.
01:13
Multiply the stoichiometric ratio from the balance equation.
01:16
You can see that for every 2 moles of mercury oxide reacted, it will produce 1 mole of o2.
01:31
2 moles of mercury oxide would produce 1 mole of o2.
01:43
Cancel grams of mercury oxide, moles of mercury oxide, and we're left with moles of oxygen gas.
01:49
59 .6 divided by 216 .59 multiplied by 2.
01:54
This will give us 0 .1376...