1. Solutions
stoichiometry

2. Stoichiometry overview
Recall that in stoichiometry the mole ratio provides a necessary conversion factor:
grams (x)  moles (x)  moles (y)  grams (y)
molar mass of x
molar mass of y
mole ratio from balanced equation
We can do something similar with solutions:
volume (x)  moles (x)  moles (y)  volume (y)
mol/L of x
mol/L of y
mole ratio from balanced equation
Read pg Try Q 1-3a.

3. Pg. 353, Question 1
Ammonium sulfate is manufactured by reacting sulfuric acid with ammonia. What concentration of sulfuric acid is needed to react with 24.4 mL of a 2.20 mol/L ammonia solution if 50.0 mL of sulfuric acid is used?
H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq)
Calculate mol H2SO4, then mol/L = mol/ L
# mol H2SO4=
2.20 mol NH3
L NH3 x
1 mol H2SO4
2 mol NH3 x
L NH3
mol H2SO4 =
mol/L = mol H2SO4 / L = mol/L

4. Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s)
Pg. 353, Question 2
Calcium hydroxide is sometimes used in water treatment plants to clarify water for residential use. Calculate the volume of mol/L calcium hydroxide solution that can be completely reacted with 25.0 mL of mol/L aluminum sulfate solution.
Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s)
# L Ca(OH)2=
L Al2(SO4)3
0.125 mol Al2(SO4)3
L Al2(SO4)3 x
3 mol Ca(OH)2
1 mol Al2(SO4)3 x
L Ca(OH)2
mol Ca(OH)2 x
= L Ca(OH)2

5. 2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq)
Pg. 353, Question 3
A chemistry teacher wants 75.0 mL of mol/L iron(Ill) chloride solution to react completely with an excess of mol/L sodium carbonate solution. What volume of sodium carbonate solution is needed?
2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq)
# L Na2CO3=
L FeCl3
0.200 mol FeCl3
L FeCl3 x
3 mol Na2CO3
2 mol FeCl3 x
L Na2CO3
0.250 mol Na2CO3 x
= L Na2CO3 = 90.0 mL Na2CO3

6. 2. Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s)
Answers
1. H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq)
Calculate mol H2SO4, then mol/L = mol/ L
# mol H2SO4=
2.20 mol NH3
L NH3 x
1 mol H2SO4
2 mol NH3 x
L NH3
mol H2SO4 =
mol/L = mol H2SO4 / L = mol/L
2. Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s)
# L Ca(OH)2=
L Al2(SO4)3
0.125 mol Al2(SO4)3
L Al2(SO4)3 x
3 mol Ca(OH)2
1 mol Al2(SO4)3 x
L Ca(OH)2
mol Ca(OH)2 x
= L Ca(OH)2

7. 3. 2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq)
Answers
3. 2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq)
# L Na2CO3=
L FeCl3
0.200 mol FeCl3
L FeCl3 x
3 mol Na2CO3
2 mol FeCl3 x
L Na2CO3
0.250 mol Na2CO3 x
= L Na2CO3 = 90.0 mL Na2CO3

8. Assignment
H2SO4 reacts with NaOH, producing water and sodium sulfate. What volume of 2.0 M H2SO4 will be required to react completely with 75 mL of 0.50 mol/L NaOH?
How many moles of Fe(OH)3 are produced when 85.0 L of iron(III) sulfate at a concentration of mol/L reacts with excess NaOH?
What mass of precipitate will be produced from the reaction of 50.0 mL of 2.50 mol/L sodium hydroxide with an excess of zinc chloride solution.

9. Assignment
a) What volume of 0.20 mol/L AgNO3 will be needed to react completely with 25.0 mL of 0.50 mol/L potassium phosphate?
b) What mass of precipitate is produced from the above reaction?
What mass of precipitate should result when L of mol/L aluminum nitrate solution is mixed with L of 1.50 mol/L sodium hydroxide solution?

10. 2. Fe2(SO4)3(aq) + 6NaOH(aq)  2Fe(OH)3(s) + 3Na2SO4(aq)
Answers
1. H2SO4(aq) + 2NaOH(aq)  2H2O + Na2SO4(aq)
# L H2SO4=
L NaOH
0.50 mol NaOH
L NaOH x
1 mol H2SO4
2 mol NaOH x
L H2SO4
2.0 mol H2SO4 x
= L = 9.4 mL
2. Fe2(SO4)3(aq) + 6NaOH(aq)  2Fe(OH)3(s) + 3Na2SO4(aq)
# mol Fe(OH)3=
0.600 mol Fe2(SO4)3
L Fe2(SO4)3 x
2 mol Fe(OH)3
1 mol Fe2(SO4)3 x
85 L Fe2(SO4)3
= 102 mol

11. # g Zn(OH)2= = 6.21 g 0.0500 L NaOH 2.50 mol NaOH L NaOH x
3. 2NaOH(aq) + ZnCl2(aq)  Zn(OH)2(s) + 2NaCl(aq)
# g Zn(OH)2=
= 6.21 g
L NaOH
2.50 mol NaOH
L NaOH x
1 mol Zn(OH)2
2 mol NaOH x
99.40 g Zn(OH)2
1 mol Zn(OH)2 x
4a. 3AgNO3(aq) + K3PO4(aq)  Ag3PO4(s) + 3KNO3(aq)
# L AgNO3 =
= L = 0.19 L
L K3PO4
0.50 mol K3PO4
L K3PO4 x
3 mol AgNO3
1 mol K3PO4 x
L AgNO3
0.20 mol AgNO3 x
4b. 3AgNO3(aq) + K3PO4(aq)  Ag3PO4(s) + 3KNO3(aq)
# g Ag3PO4=
= 5.2 g
L K3PO4
0.50 mol K3PO4
L K3PO4 x
1 mol Ag3PO4
1 mol K3PO4 x
g Ag3PO4
1 mol Ag3PO4 x

12. 5. Al(NO3)3(aq) + 3NaOH(aq)  Al(OH)3(s) + 3NaNO3(aq)
# g Al(OH)3=
L Al(NO3)3
0.500 mol Al(NO3)3
L Al(NO3)3 x
1 mol Al(OH)3
1 mol Al(NO3)3 x
77.98 g Al(OH)3
1 mol Al(OH)3 x
21.4 g Al(OH)3 =
# g Al(OH)3=
L NaOH
1.50 mol NaOH
L NaOH x
1 mol Al(OH)3
3 mol NaOH x
77.98 g Al(OH)3
1 mol Al(OH)3 x
9.36 g Al(OH)3 =
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