To fully understand the concept of limiting reactant let's make a smoothie.
Our first ingredient are strawberries. We have exactly four strawberries.
Our second ingredient is a carton of milk and we have two cartons of milk.
Now these are the ingredients or the reactants and what we're trying to make
or produce is a strawberry smoothie. So our strawberry smoothie will act as our product.
Now to make a strawberry smoothie this recipe calls for one
strawberry and one carton of milk. A limiting reactant is going to be the
reactant that is used up the fastest and it will produce the least amount of
product. So let's see which ingredient or which reactant will be used up the fastest.
Remember our limiting reactant is the reactant that runs out the fastest. So in
this case the milk is going to be our limiting reactant because it ran out the
fastest. Now what is left over is our excess reactant. So the strawberries will
be our excess reactant. Then two strawberry smoothies were produced that
is known as your theoretical yield. Whatever your limiting reactant is able
to produce is known as your theoretical yield. Let's try that example.
If 14.32g of N2 reacts with 4.21g of H2 to produce NH3 what is the limiting
reactant? This question provides us with a balanced equation. Let's move on to
step one. Step one, convert grams of each reactant to grams of product. Since we're
going from grams of each reactant our first reactant is N2 so let's plan this
out. For N2 we're gonna go from grams of N2 to then moles of N2 using our molar
mass of N2 and then from moles of N2 go to moles of NH3. In order to do this
remember this is our mole to mole ratio, let's bring back our balanced equation
and then after moles of NH3 we'll get 2g of NH3 which is what
we wanted. We want to go from grams of our reactant to grams of our product and
note we will do this for both reactants N2 and H2.
We'll go from N2 to grams of NH3. So starting with our given amount, the
14.32g of N2 which was given in the initial question, we'll align the
grams of N2 and grams of N2 remember this is going back to molar mass.
28.02 grams is the molar mass of N2. So we'll place this across from
each other putting our 1 mole of N2 on top and then
what will happen is our grams of N2 and grams of N2 will cancel. So now that our
grams of N2 cancelled out, we are on our next step which is going from moles of
N2 to moles of NH3. So since we're going from moles of one compound to moles of a
completely different compound we have to use our balanced equation. So where I'm
gonna pull these moles or this mole to mole ratio is from the coefficients or
numbers in front of our compounds in our balanced equation. So we'll align the
moles of N2 and moles of N2 across from each other. Where I got this 1 from was,
if there is no number in front of N2 or no number in front of that compound then
assume that there's a 1 in front and that's exactly what I placed here for
our mole to mole ratio. Next I'll go to moles of NH3 and we have 2 moles of NH3
going back to our balanced equation that's where I pulled that number
from we have 2 moles of NH 3 in our balanced equation. Now our moles of N2
and moles of N2 will then cancel. Now that our moles of N2 have cancelled we're
at moles of NH3, so we're at this part. Our last step here now is to go from
moles of NH3 to grams of NH3. Whenever we're going from moles to grams
we'll use the molar mass and in this case we'll use the molar mass of NH3.
So we'll align the moles of NH3 and moles of NH3 across from each other and then
we'll put our molar mass or grams on top for NH3. Now what will happen is the
moles of NH3 Will cancel and we will be left with grams of NH3 which is what we
wanted to get. We started with grams of N2 and we worked our way to grams of NH3.
So this was our grams of our reactant to grams of our product. After we multiply
straight across and then divide by 28.02 we'lll get 17.42
grams of NH3. We went from grams of N2 to grams of NH3, we have to do the same
exact process but this time for H2. So once again we'll go from grams of our
reactant to grams of our product and how we'll do this is starting with our grams
of H2 we'll convert this to moles of H2 using our molar mass of H2 then from
moles of H2 to moles of NH3 using a mole to mole ratio which comes from our
balanced equation and then finally converting our moles of NH3 to grams of
NH3 using our molar mass of NH3
So starting with our given value with
4.21g of H2, we'll align that with our molar mass of H2 so putting our
grams and our grams and then on top we'll put 1 mole of H2. Our grams of H2
would then cancel. Now that our grams of H2 have cancelled we'll move on to moles
of H2 and convert that to moles of NH3. So we'll put the moles of H2 across from
each other and where I got this 3 moles of h2 was from our balanced equation.
Since there's a 3 coefficient in our balanced equation there's really 3 moles
of H2 for every 2 moles of NH3 which is exactly what I'm using here to change
that moles of H2 to moles of NH3. Our moles of H2 will then cancel. Now that
our moles of H2 have cancelled and we're at moles of NH3 then we can finally get
to our last step which is getting to grams of NH3. So align those moles of
NH3 across from each other and put your
molar mass of NH3 on top. Our moles of NH3 would then cancel. Now that we
finally have gotten to our grams of our product from going from grams of H2 and
working our way to grams of NH3, we'll multiply straight across
and divide by 2.02 times 3 and that'll give us
23.68g of NH3. Now let's move on to step two which is our last
step. The reactant that produces the least amount of product is the limiting
reactant. And now let's compare our reactants and see which one produced the
least amount of product. N2 produced 17.42g of NH3.
And our other reactant H2 produced 23.68g of
NH3. So our N2 produced the least amount of product which was our NH3.
So finally N2 is our limiting reactant. Hey guys I created detailed notes with
examples that show every single step to find limiting reactants, excess reactants,
theoretical yield, percent yield, actual yield, all the yields. The link is in the
description box make sure to check it out. Now I know this seems like a long
process but you could do this. Back when I learned this I used to struggle with
it too but with practice and persistence you will absolutely pull through.
Alright I'll see you in the next video, I'm off to answer your comments now.