What mass of excess reactant will remain after the reaction of 35 grams of magnesium and 15 grams of nitrogen? So what I've gotta figure out is, which reagent is my limiting reagent or limiting reactant? And then figure out how much I actually need of that particular reactant. So the first thing I want to do is actually find the molar masses of each of these substances, just so I have them available when I begin to do my calculations. So I'm going to look at the periodic table and find the molar mass of magnesium is 24.31 grams per mole. For nitrogen, each nitrogen is 14.01. So my nitrogen is going to be 28.02 grams per mole. And then I can take that information and also find the molar mass of magnesium nitride. And I find that to be 100.93 grams per mole. So now I have all those numbers. So when I do need them, they're readily available to me. Now I need to take the approach of assuming that one of my reactants is limiting. And it doesn't really matter which one I start with. So we're going to look at assuming the magnesium is limiting first. So if I assume magnesium is my limiting reagent or LR, then I see that I'm going to have 35 grams of magnesium. And I'm going to have to get that to mole because remember, all my calculations need to be in terms of moles when I start to look at that stoichiometric relationship between two substances. So I have 24.31 grams per mole of magnesium. Note that I put the grams on the bottom so that those will cancel out with one another. Now I'm going to look to see which one's going to produce the least amount of product, and so here I'm going to say three moles of magnesium for every one mole of magnesium nitride. And I get those numbers from the coefficients in my balanced chemical equation. And now I'm going to actually just leave this here at moles of magnesium nitride because notice that my product, or my problem doesn't actually ask for the grams of magnesium nitride. It's asking for something else, so I'm going to just calculate the moles and I find 35 divided by 24.31, divided by my 3, that I would get 0.480 moles of magnesium nitrite. Now, I'm going to do a very similar calculation, only now, I'm going to assume that the nitrogen is limiting. So, I'm going to again start with the amount that was given of my nitrogen, which was 15.00 grams. I'm going to use my molar mass, again making sure the grams are on the bottom, so that it will cancel out. I know that for every one mole of nitrogen, I produce one mole of magnesium nitride. And again, I don't need to convert to grams because I'm really just using this calculation to determine which species is my limiting reagent. So 15 divided by 28.02, and I get 0.560 moles of magnesium nitride. This lets me know that the magnesium is actually my limiting reagent. So even though it's present in the larger amount, it's still the limiting reagent because it produces the least amount of product, and that's ultimately what I'm concerned about. What species produces the least amount of product? So now I know what my limiting reagent is. So we know this, the magnesium is limiting. So now what I need to do is actually figure out how much nitrogen do I need to react with 35 grams of magnesium? So I can say well, If I start with my 35 grams of magnesium, again use my molar mass, 24.31 grams per mole of magnesium. Now instead of looking at the relationship between magnesium and magnesium nitride, now what I'm going to look at is the relationship between magnesium and nitrogen, which is three moles of magnesium for one mole of nitrogen. Now I do need to go to grams on this one because I am trying to find the mass of the excess reactant. So I've already identified that my nitrogen is my excess reactant. And so I gotta figure out how much extra do I have? So what I'm really figuring out here is how much nitrogen do I need? If I want to completely consume all of that magnesium, how much nitrogen needs to react with that? And so now, I can use the molar mass of nitrogen. And now I can actually cancel out my units here, grams of magnesium, moles of magnesium, moles of nitrogen and moles of nitrogen. And when I do the calculation, then what I'm going to see is the grams of nitrogen needed to completely react with that 35 grams of magnesium. When I do the calculation, I get that I'm going to need 13.45 grams of nitrogen, and that's the amount of nitrogen I need to completely consume my entire 35 grams of magnesium. So what I'm trying to find out is the mass of the excess reactant. Well, I know that's the nitrogen. I know I start with 15 grams of nitrogen and I'm going to consume 13.35 grams of nitrogen in my reaction with the magnesium, and so what I find is I get 1.55 grams of nitrogen left over.
