Limiting Reagents: Movie Text

Earlier, we were working with the reaction:

Pb(NO₃)₂ + 2 NaI → PbI₂ + 2 NaNO ₃

and found that 9.05 g of NaI react completely with 10.0 g of Pb(NO₃)₂. So if we mix 9.05 g of NaI with 10.0 g of Pb(NO₃)₂, they react completely to form PbI₂ and NaNO₃. We call this "stoichiometric proportions", meaning that the amount of NaI we are adding is exactly the amount needed to react with the 10.0 g of Pb(NO₃)₂.

Let's say that I'm not being quite so careful, and instead of mixing 9.05 g of NaI with 10.0 g of Pb(NO₃)₂, I mix 10.0 g of NaI with 10.0 g of Pb(NO₃)₂. I then ask myself:

What do I have in my solution after this reaction occurs?

The reaction is going to occur until something runs out. For every molecule of Pb(NO₃)₂ that goes away, I'm going to lose two molecules of NaI. So eventually I will run out of either Pb(NO₃)₂ or NaI . If I mix together stoichiometric proportions (9.05 g NaI with 10.0 g Pb(NO₃)₂), then they both run out at the same time. But in the more general case, I need to first figure out: which reactant runs out first, Pb(NO₃)₂ or NaI?

We call the reactant that runs out first, the "limiting reagent". So a chemist would rephrase this question as:

What is the limiting reagent, Pb(NO₃)₂ or NaI?

To figure this out, we'll recast it as two separate questions.

1) How much PbI₂ would I get if Pb(NO₃)₂ is the thing that runs out first, in other words, if Pb(NO₃)₂ is the limiting reagent?

I then ask the same question for the other substance:

2) How much PbI₂ would I get if NaI is the thing that runs out first, that is, if NaI is the limiting reagent?

Whichever of the above two questions gives me the smaller number is the one that runs out first. It limits the amount of PbI₂ I can form and so it is the limiting reagent.

I'll begin with the first question. Reactions occur on the molecular scale, so I begin by converting my 10.0 g of Pb(NO₃)₂to moles of Pb(NO₃)₂. I can next use the chemical reaction to determine the number of moles of PbI₂ produced form this amount of Pb(NO₃)₂. I do this by multiplying by the ratio: (1 mol of PbI₂ produced/ 1 mol of Pb(NO₃)₂ consumed), and find that 0.0302 moles of PbI₂ are produced.

So the answer to my first question is: if Pb(NO₃)₂ is the limiting reagent, I would produce 0.0302 mol PbI₂

Now, I'll do the same for question 2 regarding NaI. I take my 10.0 g of initial NaI, and use the molecular weight to convert from grams to moles. I then look at the chemical reaction and see that for every 2 moles of NaI consumed, 1 mole of lead iodide is produced, which leads to 0.0334 moles of PbI₂ produced.

What the answer to these two questions tells me is that if I were to consume all the Pb(NO₃)₂, I would get 0.0302 moles of PbI₂. Whereas if I were to consume all of the NaI, I would get 0.0334 moles of PbI₂. Since 0.0302 is less than 0.0334, I know that the amount of PbI₂ produced is limited by the amount of Pb(NO₃)₂ I had to start with. Pb(NO₃)₂ is the limiting reagent. So all of the Pb(NO₃)₂ is consumed and this gives me 0.0302 moles of PbI₂. I can convert this to grams of PbI₂ by multiplying by the molecular weight, and I find 13.9 g PbI₂ produced by the reaction.

I could have guessed from the start that Pb(NO₃)₂ was the limiting reagent, because I knew it only took 9.05 g of NaI to react with 10.0g of Pb(NO₃)₂. All of the extra NaI I'm putting in - the difference between 10.0 g and 9.05 g - isn't going to have any Pb(NO₃)₂ to react with; it's just going to sit there as leftover NaI.

Let's consider a general method we could use to determine the amount of NaI left over after the reaction. The grams of NaI remaining is equal to the grams of NaI I had to start with, minus the grams of NaI consumed. I know that I started with 10.0 g of NaI, so I just need to determine the amount of NaI consumed by the reaction. The difference is the amount of NaI remaining. To do this, I'll start with the amount of PbI₂ produced by the reaction, 0.0304 mol PbI₂. For every mole of PbI₂ produced, I consume 2 moles of NaI, so I can multiply by the ratio: (2 mol NaI consumed/ 1 mol PbI₂ produced). I now know the amount of NaI consumed in terms of moles of NaI. I can convert this to grams by using the molecular weight, multiplying by 149.89 g NaI per mole of NaI gives 9.05 g NaI consumed by the reaction.

I now know the number of grams NaI consumed. If I subtract this from the grams of NaI I had to start with, I get 0.95 g of NaI remaining.

So I've achieved the goal of knowing what is left over after the reaction.