Section 8.3 pg. 320-324

In any chemical reaction, it is easy to run out of one or another reactant – which has an impact on the amount of products that can result from a reaction

To solve these problems you must identify which of the reactants is going to run out first. This is the “limiting reagent”The other is the “excess reagent”

Example: LEGO – If you had three yellow blocks and four red blocks, you could only make three yellow/red combinations – because there is not enough yellow blocks to make four. The yellow block is the limiting reagent and the red block is the excess reagent

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WHY DO WE CARE??

It is often desirable to know how much excess reagent is required to ensure that a reaction goes to completion.The general rule is to assume that a reasonable quantity

of excess reagent is to use 10% more than the quantity required (Not the case in commercial chemistry)

When you know the quantity of more than one reagent, you may need to know which one will limit the reaction.

1) You want to test the stoichiometric method using the reaction of 2.00 g of copper(II) sulfate in solution with an excess of sodium hydroxide in solution. What would be a reasonable mass of sodium hydroxide to use?

To answer this question, you need to calculate the minimum mass required and then add 10%.

CuSO4(aq) + 2 NaOH(aq) Cu(OH)2(s) + Na2SO4(aq)

2.00g m = ? 159.62g/mol 40.00g/mol

2.00 g x 1 mol x 2 x 40.0g = 1.00 g 10% = 0.10g 159.62 g 1 1 mol

1.00 g x 1.10%= 1.10g1.10g Practice pg. 321 #2

3) If 10.0 mol of methane and 10.0 mol of oxygen react, which is the limiting reagent.

10.0 mol 10.0 mol

n CH4(g): 10.0 mol x 2 = 20.o mol 1

n O2(g): 10.0 mol x 1 = 5.00 mol 2

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

20.0 mol of oxygen would be

required to react with 10.0 mol of methane

5.00 mol of methane would be required to react with 10.0 mol of oxygen

Since 20.0 mol of oxygen is required to react with 10.0 mol of methane, but only 10.0 mol is available, oxygen is the limiting reagent.

How much methane would be left? 5.0 mol – 5.00 mol = 5.00 mol Practice pg. 324

#3b-d

2) If 10.0g of copper is placed in solution of 20.0g of silver nitrate, which reagent will be the limiting reagent?

All reactants must be converted to moles, then using the mole ratio, determine which reactant will run out first.

Cu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq)

10.0g 20.0g

63.55 g/mol 169.88g/mol

n Cu(s): 10.0g x 1 mol = 0.157 mol x 2 = 0.315 mol = 0.315 mol

63.55 g 1

n AgNO3: 20.0g x 1 mol = 0.118 mol x 1 = 0.0589 mol= 0.0589 mol

169.88 g 2

You must test one of the values using the mole ratio. Assume one chemical is completely used up and see if enough of the second chemical is present.

That much silver nitrate is not available so copper is not the limiting reagent

More copper than that is available so silver nitrate is the limiting reagent

3) From the previous example, where 10.0 g of copper reacts with 20.0 g of silver nitrate, what mass of copper will be in excess? (leftover when the reaction is complete)

Cu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq)

0.157 mol 0.118 mol

n Cu(s): 0.118 mol x 1 = 0.0589 mol x 63.55 g = 2 1 mol

10 g – 3.74 g = 6.3 g of copper will be leftover

LRER

3.74 g of copper will be required in this reaction

4) What mass of silver will be produced? n Ag(s): 0.118 mol x 2 = 0.118 mol x 107.87

g = 2 1 mol

12.7 g of silver will be produced in this reactionPractice pg. 324

#4

Putting it all together… In an experiment, 26.8g of iron (III) chloride in solution is

combined with 21.5g of sodium hydroxide. Which reactant is in excess, and by how much? What mass of precipitate will be obtained?

FeCl3(aq) + 3NaOH(aq) Fe(OH)3(s) + 3NaCl(aq) 26.8g 21.5g m = ? 162.20g/mol 40.00g/mol 106.88g/mol

nFeCl3 = 26.8g x 1 mol = 0.165 mol x 3 = 0.496 mol 162.20g 1

nNaOH = 21.5 g x 1 mol = 0.538 mol 40.0g

0.538mol – 0.496 mol = 0.042 mol x 40.00 g = 1.7 g NaOH (excess)

1 mol

0.165 mol x 1 x 106.88 g = 17.7 g of Fe(OH)3(s) 1 mol

There is more NaOH than

this so FeCl3 is the LR

Identify the limiting reagent by choosing either reagent amount, and use the mole ratio to compare the required amount with the amount actually present.

The quantity in excess is the difference between the amount of excess reagent present and the amount required for complete reaction.

A reasonable reagent excess to use to ensure complete reaction is 10%.

Pg. 327 #4, 6-8