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## Limiting Reactant

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### Limiting Reactant

The reactant that limits the amount of product which can form.

Limiting reactant -

The reactant that is completely consumed before all of the other reactants are converted to product.

Excess Reactant –

Reactant that is not completely consumed in the reaction.

Some of this reactant will remain when the reaction is complete.

Steps for Determining the Limiting Reactant:

1. Pick a product to work with (doesn’t matter which one).

2. If the reactants are in grams change each to moles.

3. Take one of the reactants and calculate how many moles of product you can make.

4. Take the other reactant and calculate how many moles of product you can make.

5. Compare the amount of product you can make from step 2 and 3. The reactant that makes the least amount of product will be the one that limits the reaction.

Sandwich Analogy

Suppose you own a sandwich shop.

In order to make a ham sandwich, you must have 2 slices of bread and 1 slice of ham.

+

2 B

+

1 H

1 B2H

Sandwich Analogy Cont.

1. Pick a product to work with (doesn’t matter which one).

In this case, we only have one product, sandwiches.

+

2 B

+

1 H

1 B2H

2. If the reactants are in grams change each to moles. (We’ll assume we are given moles in this instance).

3. Take one of the reactants and calculate how many moles of product you can make. 54 Slices of Bread

4. Take the other reactant and calculate how many moles of product you can make. 40 Pieces of Ham

54 B

1 B2H

= 27 B2H (Sandwiches)

2 B

40H

1 B2H

= 40 B2H (Sandwiches)

1H

+

2 B

+

1 H

1 B2H

Sandwich Analogy Cont.

How many sandwiches can we make?

5. Compare the amount of product you can make from step 2 and 3. The reactant that makes the least amount of product will be the one that limits the reaction.

+

Even though we have enough ham for 40 sandwiches, we only have enough bread for 27 sandwiches.

Therefore only 27 sandwiches can be made.

Since bread LIMITS the amount of product we can make, it is the LIMITING REACTANT!

Practice Problem

If 42.0 grams of sodium phosphate are reacted with 37.6 grams of aluminum nitrate:

a. What is the limiting reactant?

b. What reactant is in excess?

c. What mass of aluminum phosphate will be precipitated?

? LR? Excess ?g AlPO4

42.0 g Na3PO4 37.6g Al(NO3)3

42.0 g Na3PO4 37.6g Al(NO3)3

Na3PO4

+

Al(NO3)3

3 NaNO3

+

AlPO4

1. Pick a product with which to work.

AlPO4

2. If the reactants are in grams change each to moles.

3. Take one of the reactants and calculate how many moles of product you can make.

4. Take the other reactant and calculate how many moles of product you can make.

42.0 g Na3PO4

1 mol Na3PO4

1 mol AlPO4

= 0.256 mol AlPO4

164.0 g Na3PO4

1 mol Na3PO4

37.6 g Al(NO3)3

1 mol Al(NO3)3

1 mol AlPO4

= 0.177 mol AlPO4

213.0 g Al(NO3)3

1 mol Al(NO3)3

5. The reactant that makes the least amount of product will be the one that limits the reaction.

? LR? Excess ?g AlPO4

42.0 g Na3PO4 37.6g Al(NO3)3Na3PO4

+

Al(NO3)3

3 NaNO3

+

AlPO4

42.0 g Na3PO4

1 mol Na3PO4

1 mol AlPO4

= 0.256 mol AlPO4

164.0 g Na3PO4

1 mol Na3PO4

37.6 g Al(NO3)3

1 mol Al(NO3)3

1 mol AlPO4

= 0.177 mol AlPO4

213.0 g Al(NO3)3

1 mol Al(NO3)3

Limiting Reactant =

Al(NO3)3

Excess Reactant =

Na3PO4

42.0 g Na3PO4 37.6g Al(NO3)3

42.0 g Na3PO4

1 mol Na3PO4

1 mol AlPO4

NOT Going to Happen!

= 0.256 mol AlPO4

164.0 g Na3PO4

1 mol Na3PO4

37.6 g Al(NO3)3

1 mol Al(NO3)3

1 mol AlPO4

= 0.177 mol AlPO4

213.0 g Al(NO3)3

1 mol Al(NO3)3

Aluminum nitrate is the limiting reactant. Only 0.177 moles aluminum phosphate can be made!

0.177 mol AlPO4

122g AlPO4

= 21.6 g AlPO4

1 mol AlPO4

Finding the amount of excess reactant used:

Take the amount of product you will be able to make (based on the limiting reactant) and with that amount, calculate the moles of excess reactant it takes to make that amount of product. This is the amount USED.

Translation:

Convert the moles of product to excess reactant.

54 B

1 B2H

= 27 B2H (Sandwiches)

2 B

40H

1 B2H

= 40 B2H (Sandwiches)

NOT Going to Happen!

1H

Take the product, 27 B2H and convert it to excess reactant, Ham.

27 B2H

1 H

= 27 H

1 B2H

This is how much ham we USED. (27H)

Finding the amount of excess reactant REMAINING:

Take the amount of excess reactant used and subtract it from the amount you started with (from the original question when you were finding limiting reactant).

Translation:

Started with

Used

=

Remaining

Sandwich Analogy Cont.

27 B2H

1 H

= 27 H

1 B2H

This is how much ham we USED. (27H)

If we started with 40 pieces of ham and used 27, how many remain?

=

Started with

Used

Remaining

40 H

27 H

=

13 H

+

Al(NO3)3

3 NaNO3

+

AlPO4

Practice Problem42.0 g Na3PO4

1 mol Na3PO4

1 mol AlPO4

NOT Going to Happen!

= 0.256 mol AlPO4

164.0 g Na3PO4

1 mol Na3PO4

37.6 g Al(NO3)3

1 mol Al(NO3)3

1 mol AlPO4

= 0.177 mol AlPO4

213.0 g Al(NO3)3

1 mol Al(NO3)3

Convert the moles of product to grams of excess reactant.

(We want grams because that was the original unit for Na3PO4, excess reactant).

0.177 mol AlPO4

1 mol Na3PO4

164.0 g Na3PO4

= 29.0g Na3PO4

1 mol AlPO4

1 mol Na3PO4

29.0g Na3PO4 was USED to make the 0.177 mol AlPO4

Practice Problem cont.

How much excess reactant remains?

42.0 g Na3PO4 37.6g Al(NO3)3

Recall our original givens:

=

Started with

Used

Remaining

42.0 g Na3PO4

29.0g Na3PO4

=

13.0g Na3PO4

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