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)
C (mol/L) of x
C (mol/L) of y
mole ratio from balanced equation

3. H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq)
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)
c = ? c = 2.20 mol/L
v = 0.05 L v = L
n = ? n = ?
(2)
(1b)
(1a)
2.20 mol NH3
L NH3 x
1 mol H2SO4
2 mol NH3 x
(n) H2SO4=
L NH3
mol H2SO4 =
(2)
Then do
c=n/v for H2SO4
n = ( V x c ) x mole ratio
(1a)
(1b)

4. Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s)
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)
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)
4 step problem! (1st being writing the balanced chemical equation)
V (FeCl3) n (FeCl3) n (Na2CO3)  V (Na2CO3)
C (FeCl3)
C (Na2CO3)
mole ratio from balanced equation

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