00:02
Question number five is a multi -step question regarding limiting reactants.
00:10
However, it's set up in such a fashion where each previous part prepares you for, in part, the next part.
00:21
The reaction, which they ask you to write first in part a, is sodium chloride reacting with silver nitrate, producing silver chloride solid and sodium nitrate.
00:34
It is balanced as written, no coefficients are required.
00:39
In part b, it asks you to determine how many moles of each reactant are you starting with.
00:44
Well, they tell us we're starting with 6 .3 grams of silver nitrate.
00:50
So all we need to do is divide that by the molar mass of silver nitrate in order to get the moles silver nitrate, 3 .71 times 10 to the negative 2.
01:01
We are also starting with 4 .5 grams.
01:05
Sodium chloride.
01:08
We can get moles sodium chloride by dividing by molar mass sodium chloride which is 7 .70 times 10 to the negative 2.
01:17
Part c asks you to determine what is the limiting reactant.
01:22
To do this appropriately you need to figure out how many moles of either product just choose one can be created with these many moles of reactant.
01:35
So we'll convert the 3 .71 times 10 to the negative 2 mole's silver nitrate into moles silver chloride, this being the product i've chosen, and we get 3 .71 times 10 to the negative 2 moles silver chloride can be made from the amount of silver nitrate we have.
01:54
Then we'll take the 7 .70 times 10 to negative 2 moles sodium chloride, and the stoichiometry here is also 1 to 1.
02:03
It's one to one for everything throughout.
02:07
And we get 7 .70 times 10 to negative 2 moles silver chloride...