stoichiometry excess-lim scott

ofExcess-Limiting

Reactions

Excess-Limiting ConceptConsider the simple reaction:

A + B CIt means that 1 mole of “A” reacts with

1 mole of “B” and produces 1 mole of “C”.

+A = B =C =

?+

But, what if we actually put 2 moles of “A” into a container with only 1 mole of “B”?

There is one excess “A” left over when the reaction is finished. Why does this happen?

EXCESS

Excess-Limiting ConceptA + B C

According to the equation, they MUST combine in a 1:1 ratio. After the 1 mole of B is all used up, we

will have only used 1 mole of A and there will still be 1 mole of A left over.

?+EXCESS

A =

B =

C =

Because there is more than enough A for the reaction, A is called the excess reactant.

When B is used up, the reaction stops. B is called the limiting reactant because it limits how

much C the reaction can make.

Excess-LimitingLet’s see how this works with an actual reaction.

Consider the reaction: 3 Mg + N2 Mg3N2

Suppose we have 10. grams of Mg and 5.0 g of N2

for this reaction. How much product will we be able to make?

10. g 5.0 g ?

We can quickly see that this reaction requires 3 moles of Mg to every 1 mole of N2.

In other words, it takes 3 times as many Mg atoms as it does N2 molecules.


Consider the reaction: 3 Mg + N2 Mg3N210. g 5.0 g ?

The other issue is molar mass. Mg has a molar mass of 24.3 g/mol and the

N2 has a molar mass of 28.0 g/mol. How does this relate?

It’s important because if we have equal masses of these two, we will actually have more Mg atoms than N2 molecules

since each Mg is lighter in mass.



We need to identify the limiting reactant… the one that will be used up first, before we can determine the amount of

product that can be made in this situation.

This is done by using the usual stoichiometric process of: grams A moles A; moles A moles B; moles B grams B



We begin by using what we “have” of each reactant and calculate the amount of each that we would “need”.

2 22

2

22

2 2

1mol N 28.0g N1mol Mg10.g Mg =3.84g N24.3g Mg 3mol Mg 1mol N

1mol N 3mol Mg 24.3g Mg5.0g N =13g Mg28.0g N 1mol N 1mol Mg

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We “have” this much

We “need” this much



2 22

2

22

2 2



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To react all of the 10.g of Mg that we “have”, we “need” 3.84g of N2.

5.0g of N2 – 3.84g N2 needed leaves 1.16 g excess N2.

We already “have” 5.0 g of N2 so since we only “need” 3.84g of N2, we have more than we need.



2 22

2

22

2 2



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5.0g of N2 – 3.84g N2 needed leaves 1.16 g excess N2.

The 1.16 g of excess N2 will not react because there is not enough Mg to react with it. Since there will be left over N2

that is un-reacted, we call the N2 the excess reactant.

Excess



2 22

2

22

2 2



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Comparing the 10. g of Mg that we “have” to the 13 g of Mg that we “need”, we see that we don’t “have” enough Mg to

react with all of the N2.

When the 10. g of Mg is used up, the reaction will stop.

Excess



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2

22

2 2



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With all of the Mg gone, no more N2 can be used and no more product can be made.

Mg is called the limiting reactant because when it is gone, the reaction stops. It therefore “limits” the amount of product

that can be made.

Excess

Limiting



2 22

2

22

2 2



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Now we can answer the original question which was how much product will we make?

Excess

Limiting

1mol Mg10.g Mg 24.3g Mg⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

Using the Limiting3 21mol Mg

3mol MgN⎛ ⎞

⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

3 23 2

100.9 Mg1mol Mg

NN

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

3 2= 13.8 g Mg N

Have we learned it yet?Try this one on your own:

Al2O3 + 6 HF → 2 AlF3 + 3 H2OHow many grams of H2O can be made with 150g of aluminum

oxide and 150g of hydrofluoric acid?

Follow these steps:• Use the 150g of Al2O3 to find the mass of HF needed.• Use the 150 g of HF to find the mass of Al2O3 needed.• Compare the amounts you “have” to the amounts you “need”

and determine the limiting reactant (have<need).• Use the limiting reactant amount to calculate the mass of H2O

that will be produced.

Al2O3 + 6 HF → 2 AlF3 + 3 H2OAnswer

2 32 3

2 3 2 3

2 3 2 32 3

1mol Al O 20.006g HF6mol HF150g Al O =177g HF101.9613g Al O 1mol Al O 1mol HF

1mol Al O 101.9613g Al O1mol HF150g HF 20.006g HF 6mol HF 1mol Al O

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⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

HAVE NEED

Excess

Limiting

2 22

2

3mol H O 18.02g H O1mol HF150g HF =127g H O20.006g HF 6mol HF 1mol H O

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stoichiometry excess-lim scott

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