Solution stoichiometry

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SOLUTION STOICHIOMETRY
Problems and solutions
VOCAB: THE BASICS FIRST
Strong Acids and Bases: Dissociate 100% in water
Acids: CBS PIN
HCl HBr H2SO4
HClO4 HI HNO3
Bases: Group IA (Li-Cs) and IIA (Mg-Ba)
hydroxides
Mg(OH)2
H2SO4
Mg2+ + 2OHH+ + HSO4-
EXAMPLE #1
DILUTION
If you need to make 75ml of 1.5M HCl solution
from 12M stock HCl. How much stock solution and
water are needed?
M1V1 = M2V2
75ml(1.5M) = (12M)(V2)
9.4ml of stock and 65.6ml H2O
EXAMPLE #2 STOICHIOMETRY
What volume of Na3PO4 (6.0M) is needed to react
with 120ml of 4.2M ZnCl2?
Both reactants are soluble:
Na3PO4
3Na+ + PO43ZnCl2
2Cl- + Zn2+
Net ionic:
3Zn2+ + 2PO43-
Zn3(PO4)2
EXAMPLE #2…CONT.
Net ionic:
3Zn2+ + 2PO43-
Zn3(PO4)2
Use stoichiometry:
.120L 4.2molZnCl2 1molZn2+ 2molPO43- 1molNa3PO4 1L =
1L
1molZnCl2 3molZn2+ 1molPO436molNa3PO4
=.056L = 56ml Na3PO4
EXAMPLE #3
A 1.42g sample of a pure compound, with the
formula M2SO4, was dissolved in water and treated
with an excess of aqueous calcium chloride, resulting
in the precipitation of all the sulfate ions as calcium
sulfate. The precipitate was collected and found to
have a mass of 1.36g.
Determine the atomic mass of M and identify M
EXAMPLE #3 – JUST ONE METHOD OF SOLVING
M2SO4 + CaCl2
1.42g
1.36gCaSO4 1molCaSO4
136g
2MCl + CaSO4
1.36g
1molM2SO4
1molCaSO4
Xg
=
1mol M2SO4
.01x g = 1.42g M2SO4
X=142g M2SO4 subtract the mass of SO4
2M = 45.94g therefore, mass M = 22.97g which is Na
EXAMPLE #4
Consider a 1.50g mixture of magnesium nitrate and
magnesium chloride. After dissolving this mixture in
water, 0.500M silver nitrate is added drop-wise until
precipitate formation is complete. The mass of the
white precipitate formed is 0.641g.


Calculate the mass percent of magnesium chloride
in the mixture
Determine the minimum volume of silver nitrate
that must be added to ensure the complete
formation of the precipitate.
EXAMPLE #4 – JUST ONE METHOD OF SOLVING
MgCl2 + 2AgNO3
1.50g mixture
0.500M
.641gAgCl
1molAgCl
143.15g
2AgCl +
Mg(NO3)2
.641g
1molMgCl2
2mol AgCl
95.21g = .213gMgCl2
1molMgCl2
.213gMgCl2 / 1.50g mixture = 14.2%
.213gMgCl2 1molMgCl2 2molAgNO3 1L
= 8.95ml AgNO3
95.21g
1molMgCl2 .500M
AgNO3
EXAMPLE #5
A 230. ml sample of 0.275M CaCl2 solution is left
on a hotplate overnight. The following morning, the
solution is 1.10M.

What volume of water evaporated from the
0.275M CaCl2 solution?
EXAMPLE #5 – JUST ONE METHOD OF SOLVING
CaCl2
.275M 230ml (.230L)
.275M = Xmol / .230L
.0633mol CaCl2
1.10M = .0633mol / XL
.0575L = vol in morning
230ml-57.5ml = 173ml evaporated
EXAMPLE #6
MIXING
Given 120ml of 2.2M LiCl and 400ml of
4.2M AlCl3, what is the concentration of Clafter mixing?
1.
Write dissociation reactions:
LiCl
AlCl3
Li+ + ClAl3+ + 3Cl-
Determine the moles of Cl- in each reaction:
.120L 2.2mol LiCl 1mol Cl- = 0.264 mol Cl1L
1mol LiCl
2.
EXAMPLE #6….CONT.
.400L 4.2mol AlCl3 3mol Cl- = 5.04mol Cl1L
1mol AlCl3
3.
4.
Add moles of Cl- together = 5.304 mol ClDivide moles by the total volume of the
solution.
5.304mol Cl- / .520L = 10.2M

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