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SOLUTION STOICHIOMETRY
- Previously, we dealt with comparing components of a chemical reaction by
converting mass of substances to moles.
- Now we will consider solutions where we will need to convert volumes to moles
to make comparisons.
Scheme:
Mass of
reactant
(g)
Mass of
product
(g)
M
Molar
mass
M
Molar
mass
Moles of
reactant
(mol)
Balanced
equation
Moles of
product
(mol)
M
Molarity
M
Molarity
Volume of
reactant
(L)
Volume of
product
(L)
GRAVIMETRIC ANALYSIS
Measure the concentration of solution by measuring the mass of a precipitate
formed.
- Use grams of precipitate to find moles of solute.
- Divide by volume of solution to find concentration.
Example: 25.00 mL of Pb(NO3)2 solution with an unknown concentration reacts
with excess aqueous Rb3AsO4. After filtering and drying, 0.0814 g of
precipitate is found. What is the concentration of the lead(II) nitrate
solution? M(Pb3(AsO4)2) = 692.2 g/mol
3 Pb(NO3)2 (aq) + 2 Rb3AsO4 (aq)  Pb3(AsO4)2 (s) + 6 RbNO3 (aq)
0.0814g Pb3 (AsO4 )2 
1molPb3 (AsO4 )2

3molPb(NO3 )2
692.2g Pb3 (AsO4 )2 1molPb3 (AsO4 )2

0.0141 molPb(NO3 )2
1

0.02500 L
L
2
VOLUMETRIC ANALYSIS
Measure the concentration of solution by measuring the volume of solution need to
find stoichiometric equivalence between reactants using a titration.
- Use volume of titrant to find moles of analyte.
- Divide moles of analyte by volume of analyte to find concentration.
- Acid/base reactions are commonly used but not always.
Example: What is the concentration of a HBr solution when 50.00 mL is titrated
with 41.88 mL of 0.176 M KOH?
KOH (aq) + HBr (aq)  H2O (l) + KBr (aq)
0.04188 L 
0.1762 mol KOH 1molHBr
0.1476 mol HBr
1



 0.1476 M HBr
L
1mol KOH 0.05000 L
1L
LIMITING REAGENT PROBLEMS
Recall the following about limiting reagent problems.
- To find limiting reactant, calculate number of moles of product formed from
each number of moles of reactant.
- Limiting reactant will yield lowest number of moles produced.
Example: How many grams of Zn(OH)2 are produced when 350 mL of 0.152 M of
ZnSO4 is mixed with 250 mL of 0.275 M of LiOH?
First write balanced equation
ZnSO4 (aq) + 2 LiOH (aq)  Zn(OH)2 (s) + Li2SO4 (aq)
Now find number of moles of each reactant.
ZnSO 4 : 0.350 L 
0152
.
mol
 0.0532 mol ZnSO 4
L
LiOH: 0.250 L 
0.275 mol
 0.0688 mol LiOH
L
Now calculate possible amount of product that each reactant can produce.
ZnSO 4 :0.0532mol ZnSO 4 
LiOH :0.0688mol LiOH 
1mol Zn ( OH) 2
1mol ZnSO 4
1mol Zn ( OH) 2
2mol LiOH
Therefore, LiOH is the limiting reagent.
0.0688mol LiOH 
 0.0532mol Zn ( OH)2
 0.0344mol Zn ( OH) 2
1mol Zn ( OH)2 99.39g Zn ( OH)2

 3.42g Zn ( OH)2
2mol LiOH
mol Zn ( OH)2
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Example: When 732 L of 1.81 M of Ag2SO4 is mixed with 1148 L of 2.07 M of
KBr,
a) kilograms of AgBr (s) formed
b) concentration of all metal ions remaining
a) Ag2SO4 (aq) + 2 KBr (aq)  2 AgBr (s) + K2SO4 (aq)
Ag 2SO4 : 732 L 
KBr : 1148 L 
1.81 molAg2SO4
L

2 molAgBr
1molAg2SO4
 2660 mol AgBr
2.07 mol KBr 1 mol AgBr

 2380 mol AgBr
L
1mol KBr
KBr is the limiting reactant.
KBr : 2380 mol AgBr 
187.772 g AgBr
molAgBr
 447000 g AgBr  447 kg AgBr
In photographic film, AgBr decomposes on exposure to light which darkens the negative.
2 AgBr(s) + h  2 Ag(s) + Br2 (l)
b) Find concentration of metal ions remaining.
The metal ions we might have in solution are Ag+ and K+.
i) Consider the concentration of K+ first.
- K+ is a spectator ion.
- It hasn’t participated in any chemical change.
Number of moles of K+ ion is same as number of moles of KBr initially.
Volume of solution is 732 L + 1148 L = 1880 L.
c K 
2380 mol K 
1880 L
 126
. M
Note that although KBr is limiting reagent, none of the K+ is used. This means
that we could be more precise by saying that Br- is the limiting reagent.
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ii) Consider the concentration of Ag+.
Once reaction is complete, most of the Ag+ ion is now part of AgBr solid. But
since Ag+ ion is not limiting reagent some of it remains.
How much remains?
First of all how much Ag2SO4 went into AgBr solid.
2380 mol KBr 
1 mol Ag2SO4
2 mol KBr
 1190 mol Ag2SO4
Therefore subtract Ag2SO4 used from Ag2SO4 total.
total
- used
remaining
1330 mol
-1190 mol
140 mol
We need to be careful about dissociation of Ag2SO4.
140 mol Ag 2SO4 
2 mol Ag 
1 mol Ag 2SO4
 280 mol Ag 
Thus concentration of Ag+ is
c Ag  
280 mol Ag 
1880 L
 015
. M