Stoichiometry overview
•
Recall that in stoichiometry the mole ratio
provides a necessary conversion factor:
molar mass of x
molar mass of y
grams (x)  moles (x)  moles (y)  grams (y)
mole ratio from balanced equation
•
We can do something similar with solutions:
mol/L of x
mol/L of y
volume (x)  moles (x)  moles (y)  volume (y)
•
mole ratio from balanced equation
Read pg. 351-353. Try Q 1-3a.
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/0.0500 L
# mol H2SO4=
0.0244 L NH3 x 2.20 mol NH3 x 1 mol H2SO4 =0.02684 mol
L NH3
2 mol NH3
H2SO4
mol/L = 0.02684 mol H2SO4 / 0.0500 L = 0.537 mol/L
Pg. 353, Question 2
Calcium hydroxide is sometimes used in water
treatment plants to clarify water for residential
use. Calculate the volume of 0.0250 mol/L
calcium hydroxide solution that can be
completely reacted with 25.0 mL of 0.125 mol/L
aluminum sulfate solution.
Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s)
# L Ca(OH)2=
0.0250 0.125 mol Al2(SO4)3 3 mol Ca(OH)2 L Ca(OH)2
x
x
x
L Al2(SO4)3 L Al2(SO4)3
1 mol Al2(SO4)3 0.0250 mol
Ca(OH)2
= 0.375 L Ca(OH)2
Pg. 353, Question 3
A chemistry teacher wants 75.0 mL of 0.200
mol/L iron(Ill) chloride solution to react
completely with an excess of 0.250 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=
0.0750 L FeCl3 0.200 mol FeCl3 3 mol Na2CO3 L Na2CO3
x
x
x
L FeCl3
2 mol FeCl3 0.250 mol
Na2CO3
= 0.0900 L Na2CO3 = 90.0 mL Na2CO3
Answers
1. H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq)
Calculate mol H2SO4, then mol/L = mol/0.0500 L
# mol H2SO4=
0.0244 L NH3 x 2.20 mol NH3 x 1 mol H2SO4 =0.02684 mol
L NH3
2 mol NH3
H2SO4
mol/L = 0.02684 mol H2SO4 / 0.0500 L = 0.537 mol/L
2. Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s)
# L Ca(OH)2=
0.0250 0.125 mol Al2(SO4)3 3 mol Ca(OH)2 L Ca(OH)2
x
x
x
L Al2(SO4)3 L Al2(SO4)3
1 mol Al2(SO4)3 0.0250 mol
Ca(OH)2
= 0.375 L Ca(OH)2
Answers
3. 2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq)
# L Na2CO3=
0.0750 L FeCl3 0.200 mol FeCl3 3 mol Na2CO3 L Na2CO3
x
x
x
L FeCl3
2 mol FeCl3 0.250 mol
Na2CO3
= 0.0900 L Na2CO3 = 90.0 mL Na2CO3
Assignment
1. 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?
2. How many moles of Fe(OH)3 are produced
when 85.0 L of iron(III) sulfate at a
concentration of 0.600 mol/L reacts with
excess NaOH?
3. 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.
Assignment
4. 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?
5. What mass of precipitate should result when
0.550 L of 0.500 mol/L aluminum nitrate
solution is mixed with 0.240 L of 1.50 mol/L
sodium hydroxide solution?
Answers
1. H2SO4(aq) + 2NaOH(aq)  2H2O + Na2SO4(aq)
# L H2SO4=
0.075 L NaOH x0.50 mol NaOH 1 mol H2SO4
L H2SO4
x
x
L NaOH
2 mol NaOH 2.0 mol H2SO4
= 0.009375 L = 9.4 mL
2. Fe2(SO4)3(aq) + 6NaOH(aq)  2Fe(OH)3(s) + 3Na2SO4(aq)
# mol Fe(OH)3=
85 L Fe2(SO4)3 x 0.600 mol Fe2(SO4)3 x 2 mol Fe(OH)3
L Fe2(SO4)3
1 mol Fe2(SO4)3
= 102 mol
3. 2NaOH(aq) + ZnCl2(aq)  Zn(OH)2(s) + 2NaCl(aq)
# g Zn(OH)2=
= 6.21 g
0.0500 x 2.50 mol NaOH 1 mol Zn(OH)2 99.40 g Zn(OH)2
x
x
L NaOH
L NaOH
2 mol NaOH 1 mol Zn(OH)2
4a. 3AgNO3(aq) + K3PO4(aq)  Ag3PO4(s) + 3KNO3(aq)
# L AgNO3 =
= 0.1875 L = 0.19 L
0.025 0.50 mol K3PO4 3 mol AgNO3
L
AgNO
3
x
x
x
L K3PO4
L K3PO4
1 mol K3PO4 0.20 mol AgNO3
4b. 3AgNO3(aq) + K3PO4(aq)  Ag3PO4(s) + 3KNO3(aq)
# g Ag3PO4=
= 5.2 g
0.025 x0.50 mol K3PO4 1 mol Ag3PO4 418.58 g Ag3PO4
x
x
L K3PO4
L K3PO4
1 mol K3PO4 1 mol Ag3PO4
5. Al(NO3)3(aq) + 3NaOH(aq)  Al(OH)3(s) + 3NaNO3(aq)
# g Al(OH)3=
0.550 0.500 mol Al(NO3)3 1 mol Al(OH)3 77.98 g Al(OH)3
x
x
x
LAl(NO3)3 L Al(NO3)3
1 mol Al(NO3)3 1 mol Al(OH)3
=
21.4 g Al(OH)3
# g Al(OH)3=
0.240 1.50 mol NaOH 1 mol Al(OH)3 77.98 g Al(OH)3
x
x
x
L NaOH
L NaOH
3 mol NaOH 1 mol Al(OH)3
=
9.36 g Al(OH)3
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