0:18

Now, we're going to go back to our diagram showing us the relationship between

Â the unknown and the known substances.

Â Only now, instead of stopping and

Â starting with moles of our substances, we're going to go one step further out and

Â look at the masses of our known and our unknown.

Â Note, that we're still going to go to moles of our known.

Â Use the coefficients of our balanced chemical equation.

Â Find the moles of B, and then convert to the mass of B.

Â This step here in the middle,

Â converting from moles of A to moles of B using our coefficients for

Â balanced chemical equation, is going to be done in every single problem.

Â 0:56

So let's look at an example of how we would do this.

Â First, we have an equation.

Â We do want to check to make sure it's balanced.

Â And what I see, is that it's not balanced.

Â Because immediately, I notice that there are two nitrogens on the right side of

Â the equation but only one on the left side.

Â So the first thing I need to do, is go through and balance the equation.

Â I'm going to make a short list to tally up what I

Â have on either side of the reaction.

Â So I have nitrogen, oxygen, and hydrogen.

Â And I have 1 nitrogen on the left, I have 3 oxygens on the left and 2 hydrogens.

Â On the right side of my equation, I have 2 nitrogens,

Â I have 4 oxygens and I have just 1 hydrogen.

Â 1:42

So, because I see oxygen in multiple substances,

Â I'm actually going to balance that last.

Â I'm going to start with my nitrogen, and I'm going to put a 2 in front of the NO2,

Â so that changes my nitrogens to two, and it also changes my oxygens to now

Â a total of five oxygens, because I have two times two is four, plus one is five.

Â And I see that it doesn't have any effect on the hydrogen.

Â 2:15

If I put a 2 in front of HNO3 to balance my hydrogens,

Â I get the 2 hydrogens on the right side.

Â I also see that it changes my nitrogens to 3 and it changes my oxygens to 7.

Â So I have, 2 hydrogens on the right, two nitrogens plus one nitrogen for

Â three nitrogens, six plus one, seven oxygens.

Â So, now I see that again my nitrogens are out of balance.

Â I change that to a three for the nitrogen.

Â Now I have six, seven high oxygens, I still have

Â 2 hydrogens on the left and now it appears that my equation is balanced, but

Â I'm always going to go back through and double check to make sure it's correct.

Â So, we have three nitrogens on the left.

Â We have 3 nitrogens on the right.

Â We have six plus one, seven oxygens on the left.

Â We have six plus one equals seven oxygens on the right,

Â we have two hydrogens on the left, and we have two hydrogens on the right.

Â So, now that I know I have a balanced chemical equation,

Â I can proceed with the stoichiometry calculation.

Â In this case it's asking how many grams of nitric acid

Â are produced from 100 grams of nitrogen dioxide with excess water.

Â So the first thing I want to do,

Â is pull out the information that seems to be useful in this problem.

Â 3:56

And the only reason we have to worry about it being excess water,

Â is to know that we're not going to run out of the water.

Â The NO2 will be able to react completely to form the maximum amount of HNO3.

Â Now, I see that I'm going to have to convert between grams and

Â moles of substances, so I want to find the molar mass.

Â And I can do this with the information on the periodic table.

Â Looking at the molar mass of nitrogen and oxygen, considering that I

Â have two oxygens, I find the molar mass of NO2 is 46.01 grams per mole.

Â 4:29

And that, for HNO3, the molar mass is 63.01 grams per mole.

Â Note, that I find the molar mass for

Â the substance as written, excluding any coefficients.

Â I'll take the coefficients into account when I do my calculation.

Â But the molar mass of a substance is going to be the same regardless of

Â what its coefficient is or isn't in a chemical equation.

Â 4:52

Now, I look and see which of these but,

Â substances will allow me to find the moles of that substance.

Â If I look at HNO3, I don't know the mass, I only know the molar mass.

Â So, I'm not going to be able to determine the moles of HNO3.

Â So, that's not going to be my starting point.

Â For water, I know we have excess water, but

Â I don't know a specific amount of water.

Â So, that's not going to help me get to moles of some known quantity.

Â 5:18

The only thing that remains is my NO2.

Â I have 100 grams of NO2.

Â Now, given the molar mass of NO2,

Â I can actually go from grams of NO2 to moles of NO2.

Â And so I use the molar mass,

Â putting the 46.01 grams on the bottom, so that my units will cancel out.

Â So now, I can cancel out grams of NO2 with grams of NO2.

Â If I stop my calculation at this point, I would have mols of NO2.

Â Which tells me how much I started with, but doesn't tell me anything about

Â the mass of HNO3, which is what is being asked for in the question.

Â So now, I need to use my mol to mol ratio.

Â Just as we did in the previous calculation,

Â where we were going from moles of one substance to moles of another.

Â I still have used by mass to get to moles.

Â Now I can put in my mole ratio.

Â Now I'm going to use the coefficients from my balanced equation, so I have 3 moles

Â of NO2, from the coefficient of 3 in my equation, on the bottom.

Â And I put that on the bottom, so

Â that moles of NO2 will cancel with moles of NO2.

Â On the top, I'm going to have 2 moles of HNO3,

Â because that's what I'm trying to find.

Â So now, moles of NO2 cancels with moles of NO2.

Â And if I stopped here, I'd have moles of HNO3 as my answer.

Â But I don't want moles of HNO3, I want grams of nitric acid or grams of HNO3.

Â So, I need to do one more step to get from moles of HNO3 to grams of HNO3.

Â And again, I'm going to use my molar mass, this time for the HNO3.

Â And I know 63.01 grams per mole of HNO3.

Â Now, my moles of HNO3 cancels with my moles of HNO3.

Â I check back over my units and

Â I see that the only units I have left remaining are grams of HNO3.

Â 7:16

Now, I can do the calculation by taking

Â 100 times two times 63.01 divided

Â by 46.01 divided by three, and

Â what I find is that I get 91.30 grams of HNO3.

Â I notice that my answers are on the same order of magnitude,

Â 100 grams of NO2 to 91.3 grams of HNO3, so that seems to be a reasonable answer.

Â 7:57

Now let's look at an example problem involving the decomposition of

Â sodium azide, which is used in airbags, and

Â it produces as large volume of gas in a very short time frame.

Â Let's look at our plan to figure out how we

Â can get the mass of N2 produced from a given mass of sodium azide, or NAN3.

Â 8:17

Remember, that the only relationship we know between amounts of

Â two different substances in an equation are the molar amounts.

Â So, if I want to look at a relationship between two substances,

Â I first have to convert to moles.

Â Because if I know the moles of one substance, I can use my mole ratio for

Â my balanced chemical equation to find the moles of the other substance, and

Â then I can convert to grams.

Â 8:54

Remember, that nitrogen exists as a diatomic or N2.

Â Therefore, when I'm looking for my molar mass of N2,

Â I'm going to want to put it in there as 28.02 grams per mole.

Â If we were referring to atomic nitrogen, we would need to specify that.

Â Remember, for diatomic substances such as nitrogen or hydrogen, oxygen, when we

Â say the name of the element we assume that it is in the form of the diatomic element.

Â 9:33

We find that we can produce 6.47 grams of nitrogen.

Â Remember, that we're given our 10 grams of sodium azide, and

Â we're asked to find the grams of nitrogen.

Â Here we can set up our 10.0 grams of NaN3.

Â We need the molar mass of sodium azide, so

Â we use the values on the periodic table, and

Â that gets us 65.01 grams per mole of NaN3.

Â So now, our grams cancels with grams.

Â Now we have moles of sodium azide.

Â And now we need to use our mole to mole ratio from our balanced equation.

Â Notice, that our coefficient tells us, we have 2 moles of sodium azide and

Â we have 3 moles of Nitrogen.

Â So, our moles of sodium azide will cancel.

Â Now we have moles of nitrogen, and we're going to use our molar mass of 28.02 grams

Â per mole of nitrogen to find the grams of nitrogen.

Â And so what we find, is that we get 6.47 grams.

Â Note, that I had to convert to moles first, before I could use my mole ratio.

Â These numbers come from the balanced chemical equation.

Â