0:02 welcome everyone to the next video in
0:03 our last one we learned how to write
0:06 stoichiometric factors and in this video
0:07 we're going to learn how to use them
0:09 we're going to use them to perform
0:11 stoichiometric calculations involving
0:15 Mass moles and solution molarity
0:17 so let's do some examples
0:19 in this question we're asked how many
0:22 moles of iodine I2 are required to react
0:26 with 0.429 moles of aluminum according
0:28 to the following equation and we're
0:29 given a balanced chemical equation
0:33 involving aluminum iodine and those two
0:36 react together to form aluminum iodide
0:38 and the figure at the bottom shows
0:41 pictures of this chemical reaction where
0:43 we have the reaction of aluminum and
0:45 iodine and you can see that the heat of
0:48 the reaction actually vaporizes some of
0:51 the solid iodine to generate its purple
0:53 Vapor which is kind of cool
0:56 so how do we go about doing this
0:58 calculation well we're going to use our
1:00 stoichiometric Factor we know that our
1:03 stoichiometric Factor can relate the
1:06 number of moles of aluminum and the
1:09 moles of iodine according to the
1:11 balanced chemical equation we know that
1:14 we can write a stoichiometric factor
1:16 from our balanced chemical equation
1:19 using the ratios of these two reactants
1:22 specifically from our coefficients we
1:27 can write that three moles of I2
1:31 are needed to react with two moles of aluminum
1:33 aluminum
1:34 and I'm getting that information from
1:36 the coefficients of our balanced
1:37 chemical equation
1:40 so we know in our question that we want
1:41 to react
1:50 and we want to use this information to
1:52 calculate the moles of iodine and to get
1:54 it we're going to use our stoichiometric
1:57 Factor now we see that moles of aluminum
1:59 need to cancel so we're going to need to
2:02 write that part of our stoichiometric
2:05 factor in the denominator so we'll have
2:07 two moles of aluminum in the denominator
2:09 and in the numerator we should have
2:11 three moles
2:13 of iodine I2
2:16 when we do this calculation the moles of
2:19 aluminum should cancel and we'll be left
2:21 with moles
2:24 of I2
2:26 and plugging these numbers into a
2:27 calculator I get
2:32 0.644 so this would be 0.644 moles of I2
2:35 would be needed to react with 0.429
2:38 moles of aluminum
2:39 so now let's do an example that's
2:42 slightly more complicated so here we
2:44 have how many carbon dioxide molecules
2:47 are produced when
2:51 0.75 moles of propane is combusted
2:53 according to this equation so here we
2:55 have a combustion reaction which we've
2:57 learned here we see that our fuel
3:00 propane is going to be burned in the
3:02 presence of oxygen to yield carbon
3:06 dioxide and water now our question is
3:09 specifically asking us how many carbon
3:11 dioxide molecules are going to be
3:15 produced when 0.75 moles of propane our
3:17 fuel are combusted
3:20 now we know that we can relate moles of
3:24 propane to moles of carbon dioxide using
3:26 our stoichiometric Factor
3:28 our question is asking specifically how
3:32 many carbon dioxide molecules so once we
3:34 figure out how many moles of carbon
3:37 dioxide would be produced we need to
3:39 convert that from moles of carbon
3:42 dioxide to molecules and to do that we
3:45 can use Avogadro's number so let's do
3:46 our calculation
3:52 we have 0.75 moles of propane so that is c3h8
4:00 and to get from moles of propane to
4:02 moles of carbon dioxide I'm going to use
4:05 my stoichiometric factor I can see that
4:08 I want moles of propane to cancel so I
4:11 will need to put that in my denominator
4:14 right so that these cancel and my
4:17 stoichiometric Factor should have right
4:22 one mole of propane for every three
4:24 right moles
4:28 of CO2 carbon dioxide as I learned by
4:30 looking at the reaction Stoichiometry so
4:33 moles of propane should cancel and I'll
4:36 be left with moles of carbon dioxide
4:39 now my next step is to go from moles of
4:41 carbon dioxide to molecules of carbon
4:44 dioxide and my figure here tells me that
4:47 I can use Avogadro's number
4:48 in this case
4:51 I'm going to want moles to cancel so
4:53 I'll write one mole
4:57 of CO2 in my denominator and in my
4:59 numerator I'll use Avogadro's number so
5:05 6.022 times 10 to the 23rd molecules
5:16 and so now moles of CO2 cancel and my
5:18 calculation should give me molecules of
5:22 CO2 So my answer is 1.4 times 10 to the
5:25 24th molecules of CO2 so this tells me
5:27 how many molecules will be generated
5:32 when we combust 0.75 moles of propane
5:34 so next we'll look at more complex
