# L5 Limiting Reagent

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```					Limiting Reagent
Example

A student burns 1.2 g of butane in 1000 L
of oxygen gas at 20°C and 100 kPa. What
mass of water vapor is produced?
What is the given?
Butane and oxygen are both given.
Which one is used?
Limiting reagent

…is   the given that will be consumed first.
When the reaction consumes all of the
limiting reagent, the reaction stops.
The limiting reagent is the given to use.
Excess reagent

...is the reactant for which some will
remain unreacted after the reaction is
finished.
Sometimes a question will ask how much
unreacted of the excess reagent remains
after a reaction.

4
Example again

A student burns 1.2 g of butane in 1000 L
of oxygen gas at 20°C and 100 kPa. What
mass of water vapor is produced?
Which is the limiting reagent?
The reaction will run out of butane long
before oxygen.
Butane is the limiting reagent.
Oxygen is the excess reagent.
Butane is the given to use to answer this
question.
Example

A student   places 1.5 g of zinc into 300 mL
of 0.250 mol/L hydrochloric acid. What
mass of gas is produced?
Two givens: Zn and HCl(aq)
It is not obvious which will be consumed
first: no obvious limiting reagent.
Calculate   the answer twice: once for each
given.
The correct answer will be the smallest of
Example again

A student places 1.5 g of zinc into 300 mL
of 0.250 mol/L hydrochloric acid. What
mass of gas is produced?
Use Zn as given
Zn(s) + 2HCl(aq) → ZnCl2 (aq) + H 2 (g)
G                                 R
m       1.5 g
n=     =
M   65.41 g / mol

= 0.022932… mol of Zn
R
nR =  • nG
G
1
= • 0.022932... mol
1
= 0.022932… mol of H2

m = nM
= 0.022932… mol • 2.02 g/mol
= 0.04632319… g
= 0.0463 g = 46.3 mg
Use HCl as given
Zn(s) + 2HCl(aq) → ZnCl2 (aq) + H 2 (g)
G                        R

n = Cv = 0.250 mol/L • 0.300 L
= 0.075 mol of HCl
R
nR =   • nG
G
1
= • 0.075 mol
2
= 0.0375 mol of H2

m = nM
= 0.0375 mol • 2.02 g/mol
= 0.07575 g
= 0.0758 g = 75.8 mg
Summary

Using  Zn: 46.3 mg of H2(g) produced
Using HCl: 75.8 mg of H2(g) produced
Zn is the limiting reagent
The answer: 46.3 mg of H2(g) will be
produced.
Example

A student   dissolves 15.6 g of potassium
iodide in a solution, and 10.2 g of lead (II)
nitrate in a different solution. She mixes
the two solutions and a ppt is produced.
Identify the limiting reagent.
What mass of excess reagent will remain
unreacted?

14
Limiting reagent

To simply identify the limiting reagent,
assume one of the products is the

15
Use KI as given
2KI(aq) + Pb(NO3 )(aq) →
G                   PbI 2 (s) + 2KNO3 (aq)
R
m    15.6 g
n=     =
M 166.00 g / mol

= 0.093975… mol of KI
R
nR =  • nG
G
1
= • 0.09397... mol
2
= 0.046987… mol of PbI2

m = nM
= 0.046987… mol • 461.00 g/mol
= 21.6614… g
= 21.7 g
Use Pb(NO3)2 as given
2KI(aq) + Pb(NO3 )(aq) →
G      PbI 2 (s) + 2KNO3 (aq)
R
m      10.2 g
n=     =
M   331.22 g / mol

= 0.030795… mol of Pb(NO3)2
R
nR =  • nG
G
1
= • 0.030795... mol
1
= 0.030795… mol of PbI2

m = nM
= 0.030795… mol • 461.00 g/mol
= 14.1966… g
= 14.2 g
Limiting reagent

If KI is given, mass of ppt = 21.7 g
If Pb(NO3)2 is given, mass of ppt = 14.2 g
Pb(NO3)2 is limiting reagent.

20
Excess reagent

The  excess reagent will be the reactant
that is not the limiting reagent.
KI is the excess reagent.

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Mass unreacted

Excess  reagent, KI, is the required.
Limiting reagent, Pb(NO3)2, is the given.
Find mass of excess reagent that did
react.
Subtract this mass from initial mass.

22
Mass of excess reacted
2KI(aq) + Pb(NO3 )(aq) →
R            G      PbI 2 (s) + 2KNO3 (aq)

m      10.2 g
n=     =
M   331.22 g / mol

= 0.030795… mol of Pb(NO3)2
R
nR =  • nG
G
2
= • 0.030795... mol
1
= 0.061590… mol of PbI2

m = nM
= 0.061590… mol • 166.00 g/mol
= 10.2240… g
= 10.2 g
Excess reagent unreacted
Initial Mass
- Reacted Mass
Unreacted Mass

15.6 g
- 10.22... g
5.37597... g
5.38 g
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Page   324: #4, a) and c)

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