Checklist time:
- Write the chemical formulas for compounds
- Write chemical equations for different chemical reactions
- Calculate moles and molarity
Knowing all these things, we can now calculate theoretical yields and limiting reagents. Well, to be honest, you kinda have to calculate both if you want to know one.
Down to business. This gets a little bit more tricky than what we’ve been through so far. It isn’t anymore difficult, the process is just a bit longer.
Let’s work through an example so that it makes a teensy bit of a smidge more sense.
First things first, write down the chemical reaction. What are you waiting for? Oh right, you haven’t been given it yet. Okay, lets see, hmm…I’ve got it! Its a fun reaction that Year 7 absolutely adore! We’re going to add sulfuric acid to zinc metal. Go ahead and write it down. Done?
H2SO4 + Zn → ZnSO4 + H2
We’ll add an excess of sulfuric acid, just to make sure that all the zinc gets used up *claps hands over mouth*, whoops, I’ve just given you the answer to my next question about the limiting reagent. I’ll go through it anyways, just to be thorough.
Limiting reagent. In simple terms – the limiting reagent is just what we have the least of, it means that it limits the production of the products to a certain amount. Since we have added an excess of acid, it means that the zinc has to be our limiting reagent, its not like its self-renewing or anything (and now that just reminds me of that Tim Tam ad, mmm…chocolate). Having an excess actually makes it easier because we can now ignore the acid.
Alright, we’ve done our scientist thing and added the acid to a 5 gram strip of zinc metal. It has made pretty bubbles but now we want to know how much zinc sulfate we made.
First step?
That’s right, figure out how much zinc we had at the start (remembering that any chemical type calculations deal with moles).
Moles of zinc = 5 (g)/65.38 (g/mol)
Moles of zinc = 0.0764 mol
So, do you remember when I said that the stoichiometry of a chemical equation was important later? Well, its later now.
Basically, stoichiometry is a ratio. This example is nice and simple because it is 1:1. What that means is that 1 mole of sulfuric acid reacts with 1 mol of zinc to form 1 mol of zinc (II) sulfate and 1 mol of hydrogen gas. Except, that we don’t have 1 mol, we have 0.0764 moles. Which is fine.
To simplify, for a 1:1 ratio, its just moles of zinc = moles of zinc sulfate.
With that in mind, we can say that the moles of zinc sulfate formed = o.0764 moles.
And now all that’s left is to convert moles back into grams. We re-arrange our equation (moles=mass/molar mass) to calculate mass.
mass = moles x molar mass but remember to use the molar mass of zinc sulfate now.
mass = 0.0764 (mol) x 161.45 (g/mol)
mass = 12.33 g
Cools! We made nearly 13 grams of zinc sulfate. Well, theoretically we did. A theoretical yield is just how much you calculate that you should make. In labs, you usually calculate a % yield, which is just however much you actually made, divided by the theoretical amount.
So, if we only made 11.51 grams, our percentage yield is about 93%. Not bad!!
Just a quick note about stoichiometry. 1:1 ratios are easy-peasy. Throw in some different numbers and sometimes, its a bit hard to wrap your heads around. I won’t go through this next one completely but hopefully, you guys can.
We’ll add hydrochoric acid to lead (II) nitrate (not that they let us play with lead all that much anymore. Sad face).
2HCl + Pb(NO3)2 → PbCl2 +2HNO3
Assuming the equation is balanced (which it is but you should always check, you never know when we teachers will pull a fast one), we can see that we need twice as many moles of hydrochloric acid than lead (II) nitrate for it to react completely. So if I gave you a problem that said we had 3 moles of lead (II) nitrate and 5 moles of hydrochloric acid, which one would you say is the limiting reagent?
Yep, you got it, hydrochloric acid. So we’d only make 5 moles of lead (II) chloride at the very most.
Make sense?
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