# Topic 8.3 Stoichiometry Limiting and Excess Reagent Calculations by dfhdhdhdhjr

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```									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
   Completely consumed in a chemical reaction
   Excess Reagents
   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?

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
   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?
FeCl3(aq) + 3NaOH(aq)               Fe(OH)3(s) + 3NaCl(aq)
26.8 g     21.5g                   m            m
162.20g/mol 40.00 g/mol       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
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
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)           Cu(OH)2(aq) + Na2SO4(aq)

2.00 g           m
159.62 g/mol      40.00 g/mol
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