How many grams of magnesium do we need to start with to produce 95 grams of magnesium nitride? How many grams of nitrogen do we need? Well the first thing I notice when I look at this problem is that there are no coefficients in my equation. And that's usually a really good sign that I at least need to check to make sure I have a balanced equation. And what I see is that my nitrogens are balanced cause I have two on either side but my magnesiums are not. So I'm actually going to need to add a coefficient here, add a three in there, in order to balance the magnesium. So now I have a balanced chemical equation. The next thing I'm going to do is actually put in my molar masses, just so when I start to do my calculations I already have this information compiled. Remember that nitrogen is N2. So I have to put 28.02 grams per mole, and then magnesium nitrite is 100.93 grams per mole. So it's asking first about the amount of magnesium I need to produce some amount of magnesium nitride. So I'm going to start with my 95.00 grams of Mg3N2 and I'm going to convert to moles using the molar mass of magnesium nitride. And then I'm going to use my mole to mole ratio. So I'm talking about magnesiums, so I notice that I have one mole of magnesium nitride for every three moles of magnesium. And then I can use the molar mass of magnesium, 24.31 grams per mole of magnesium to find the grams. Now a couple things to note here, one is that I'm going to cancel my grams of magnesium nitride, moles of magnesium nitride, moles of magnesium and I'm going to be left with grams of magnesium. The other thing to note is that I started with the amount of product. And it really doesn't matter which substance I start with and which substance I'm going to. Remember that those stoichiometric coefficients from our balanced chemical equation will apply regardless of whether I'm looking at a reactant in the product, two reactants, two products or if I'm given information about the product and asking information about the reactant. It does not matter as long as I use the coefficients for my balanced chemical equation. So now I need to do the calculation, and what I find is 68.65 grams of magnesium are needed in order to produce the 95 grams of magnesium nitride. Now I need to do a similar calculation going from the magnesium nitride to the amount of nitrogen I need. So here we have, again, 95 grams of magnesium nitride. And I'm going to say use my molar mass here, 100.93 grams per mole of magnesium nitride. And now again I have to have a mole ratio. This time it's one mole of magnesium nitride to one mole of nitrogen. And then I'm going to use my molar mass of nitrogen, which is 28.02 grams per mole of nitrogen. Note that my units cancel out, my grams of magnesium nitride, moles of magnesium nitride, moles of nitrogen, and I'm left with grams of nitrogen. And I find that I need 26.37 grams of nitrogen. Now this is not a limiting reagent problem because I wasn't given two amounts of reactant. I was given a single product. And so what we've actually calculated here are the stoichiometric amounts. If I were to mix exactly 68.65 grams of magnesium with 26.37 grams of nitrogen, what I would find is that both of the reactants would be completely consumed. It would not have an excess or a limiting reagent, because both of these amounts were calculated based on producing the 95 grams of magnesium nitride.
