Limiting Reagents

Limiting Reactants/Reagents:

The reactant that is the first to run out in a chemical reaction.

Before we dive into the chemistry, let's start with a simple example:

If the recipe for an ice cream is

1 cone + 2 scoops = 1 ice cream

and I have

5 cones

12 scoops

What runs out first?

Let's try each scenario:

If we have 5 cones, we can make

(1*5) cones + (2*5) scoops = (1*5) ice creams

(since we have 5 cones, it means we can make 5x more ice cream, because the ratio is 1:2:1. Multiply that by five and you will get 5:10:5, which means we can make 5 ice creams with 10 scoops and 5 cones)

Thus we can make 5 ice creams, and that is going to need 10 scoops. Do we have have 10 scoops? Yep! We got 12, which is more than enough.

What about the next scenario?

If we use 12 scoops, we can make

(1*6) cones + (2*6) scoops = (1*6) ice creams

6 cones + 12 scoops = 6 ice creams

(Remember the ratio: 1:2:1? Our current ratio is x:12:x. How many do we multiply the ratio by? 12/2=6 We should multiply by 6. Thus our ratio is 6:12:6)

If we have 12 scoops, we can make 6 ice creams with 6 cones. (We multiply each quantity by 6, not 12, because we have 12 scoops, and each ice cream requires 2 scoops to make, so 12/2=6 and that's why we multiply the whole recipe by 6.)

However, do we have 6 cones? No! We only have 5, which means we will run out of cones first. This means cones are the limiting reagent in this case.

So how do we do this in chemistry?

Here's an example.

2Mg + O2 = 2MgO

And we have:

2.4g of Mg

10g of O

Don't panic!

1) Convert everything to moles

In this case, the mole ratio, or the numbers in the stoichiometric coefficient, is equivalent to the quantities in our ice cream recipe.

So here we go:

Mass/Molar Mass = Moles

2.4/24 = 0.1 mol of Mg

10/16 = 0.625 mol of O2

Let's stick it into each scenario:

0.1 mol of Mg

2Mg + O2 = 2MgO

|

| *0.05

\/

0.1Mg + 0.05 O2 = 0.1 MgO

So we can produce 0.1 moles of Mgo if we use 0.1 mol of Mg and 0.05 mol of O2. Do we have o.o5 mol of O2? Yes! We have 0.625 - which is in excess.

Here's the second scenario:

(We've got 0.625 mol of O2)

2Mg + O2 = 2MgO

|

| *0.625

\/

1.25 Mg + 0.625 O2 = 1.25 MgO

So with 0.625 moles of O2 and 1.25 moles of Mg, we can make 1.2 moles of MgO. However, since we don't have (or in other words will run out first) 1.25 moles of Mg, Mg is the limiting reagent. Here are the general steps: