Topic 8.3 Stoichiometry:
Limiting and Excess
Reagent Calculations
Page 320-327
By Kirsten
What is Stoichiometry?
Lets take a closer look.


A method of predicting
or analyzing the
quantities of the
reactants and products
participating in a
chemical process
Three forms including
gas stoichiometry,
solution stoichiometry,
and gravimetric
stoichiometry.
Example of Stoichiometry

In a precipitation reaction, KOH(aq)
reacts with excess Sn(NO3)2(aq) to
produce a precipitate. If the mass of
precipitate is 2.57 g, what mass of
KOH(s) was present in the original
solution?
First Step:
Balance Equation:
2KOH(aq) + Sn(NO3) 2(aq)
Sn(OH)2(s) + 2KNO3(aq)
Moles
n2
n1
Mass
?
2.57g
Molar Mass
56.11 g/mol
152.71 g/mol
Second Step:
n1= 2.57 g x (1 mol/ 152.71 g)
= 0.0168 mol
n2= 0.0168 mol x (2/1)
= 0.0337 mol
m= 0.0337 mol x (56.11 mol/g)
= 1.89g
Understanding Chemical
Principles

Conservation of Mass in a Chemical Reaction

In chemical reactions the mass is conserved,
meaning that the mass of the products equals the
mass of the reactants.


Mass Products = Mass Reactants
There will always be an equal number of moles of
each element before and after a reaction takes
place
Identifying Reagents

Limiting Reagents


Excess Reagents


Completely consumed in a chemical reaction
More is present than is necessary to react with
the limiting reagent
The limited reagent is the reagent that is
being analyzed in a quantitative analysis
where limiting and excess reagents are
present.
How to find Limiting Regent:



Ensure that all masses given are in moles.
Take the molar amounts from each
substance and divide by the coefficient of that
substance in the balanced equation.
The smaller number will be the limiting
reagent.
Example 1:

300 mL of 0.100 mol/L of BaCl2(aq) and 200
mL of 0.110 mol/L of Na2CO3(aq) are mixed.
What is the limiting reagent in the reaction?
Answer: Na2CO3
Why?
BaCl2(aq) + Na2CO3(aq)
BaCO3(s) + 2NaCl(aq)
With a 1:1 mole ratio of reactants he species
present in least amount is the limiting
reagent.
nBaCl2= 300mL x (0.100mol/L)
=30.0 mmol
nNa2CO3= 200 mL x (0.110mol/L)
= 22.0 mmol
Therefore Na2CO3(aq) is the limiting reagent
Example 2:

100.0 g of iron (III) chloride and 50.00g of
hydrogen sulfide react. What is the limiting
reagent?
2FeCl3 + 3H2S
Fe2S3 + 6HCl
Answer:



Iron(III) chloride
100g/ 162.204 g/mol
=0.6165 mol
Hydrogen Sulfide
50.00g/ 34.081 g/mol
=1.467 mol
1.467 mol/ 3
=0.489
Iron (III) chloride is the limiting reagent
Example 3:

In an experiment, 26.8 g of iron(III) chloride
in solution is combined with 21.5 g of sodium
hydroxide in solution. Which reactant is in
excess, and by how much? What mass of
each product will be obtained?
Answer:
FeCl3(aq) + 3NaOH(aq)
26.8 g
21.5g
162.20g/mol 40.00 g/mol
Fe(OH)3(s) + 3NaCl(aq)
m
m
106.88 g/mol 58.44 g/mol
nFeCl3 = 26.8 g x (1mol/ 162.20g0
=0.165 mol
nNaOH = 21.5 g x (1 mol / 40.00g)
= 0.538 mol
Answer:
nNaOH=0.165 mol x (3/1)
=0.496
nNaOH= 0.538 mol - 0.496 mol
= 0.042 mol
mNaOH = 0.042 mol x (40.00g/1 mol)
= 1.7g
Answer:
mFe(OH)3= 0.165 molFeCl3 x (1 mol Fe(OH)3/1
molFeCl3) x (106.88 gFe(OH)3/ 1 mol Fe(OH)3)
= 17.7 g Fe(OH)3
m NaCl = 0.165 molFeCl3 x (3 mol NaCl/ 1 mol FeCl3)
x (58.44 g NaCl / 1 mol NaCl)
= 29.0 NaCl
Therefore sodium hydroxide is in excess by 1.7
g, the mass of iron (III) hydroxide produced is
17.7 g and the mass of sodium chloride
produced is 29.0g.
Theoretical yields
Vs.
Actual yields

The theoretical yield of a chemical reaction is
the amount of product formed if all of the
limiting reagent reacts.



Calculated using stoichiometry.
The actual yield is the actual quantity of
products formed after a chemical reaction.
Usually the theoretical yield will be greater
than the actual yield that is produced.
Reasons for Discrepancy

The actual yield of a chemical reaction is
usually less than the theoretical yield for
these reasons



Purity of chemicals being used
Errors in measurements
Experimental factors that may have lead to loss of
reactants
% Error Calculations:

A reasonable quantity of reasonable excess
reagent is 10%
Example 4:

You decide to test the method of
stoichiometry 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?
CuSO4(aq) + 2NaOH(aq)
2.00 g
159.62 g/mol
m
40.00 g/mol
Cu(OH)2(aq) + Na2SO4(aq)
Answer:
nCuSO4 = 2.00g x (1 mol/159.62 g)
=0.0125 mol
nNaOH =0.0125 mol x (2/1)
=0.0251 mol
mNaOH = 0.0251 mol x (40g/1mol)
= 1.00 g
Now add 10% to this amount to determine the
reasonable mass of sodium hydroxide to use.
1.00 g + 0.10g
=1.10g
For More Information:

Nelson Chemistry Text



Pages 320-327
The Key- Chemistry 20
D2L Web Lessons