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. 351-353. Try Q 1-3a.

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 M ammonia solution if 50.0 mL of sulfuric acid is used?

1 mol H2SO4

2 mol NH3 x

# mol H2SO4=0.0244 L NH3 0.02684 mol

H2SO4

= 2.20 mol NH3 L NH3

x

H2SO4(aq) + 2NH3(aq) (NH4)2SO4(aq)

Calculate mol H2SO4, then mol/L = mol/0.0500 L

mol/L = 0.02684 mol H2SO4 / 0.0500 L = 0.537 mol/L

Calcium hydroxide is sometimes used in water treatment plants to clarify water for residential use. Calculate the volume of 0.0250 M calcium hydroxide solution that can be completely reacted with 25.0 mL of 0.125 M aluminum sulfate solution.

3 mol Ca(OH)2

1 mol Al2(SO4)3 x

# L Ca(OH)2=0.0250

L Al2(SO4)3

= 0.375 L Ca(OH)2

0.125 mol Al2(SO4)3 L Al2(SO4)3

x

Al2(SO4)3(aq) + 3Ca(OH)2(aq) 2Al(OH)3(s) + 3CaSO4(s)

L Ca(OH)2

0.0250 mol Ca(OH)2

x

A chemistry teacher wants 75.0 mL of 0.200 M iron(Ill) chloride solution to react completely with an excess of 0.250 M sodium carbonate solution. What volume of sodium carbonate solution is needed?

3 mol Na2CO3

2 mol FeCl3

x

# L Na2CO3=

0.0750 L FeCl3

= 0.0900 L Na2CO3 = 90.0 mL Na2CO3

0.200 mol FeCl3

L FeCl3

x

2FeCl3(aq) + 3Na2CO3(aq) Fe2(CO3)3(s) + 6NaCl(aq)

L Na2CO3

0.250 mol Na2CO3

x

Answers

1 mol H2SO4

2 mol NH3 x

# mol H2SO4=0.0244 L NH3 0.02684 mol

H2SO4

= 2.20 mol NH3 L NH3

x

1. H2SO4(aq) + 2NH3(aq) (NH4)2SO4(aq)

Calculate mol H2SO4, then mol/L = mol/0.0500 L

mol/L = 0.02684 mol H2SO4 / 0.0500 L = 0.537 mol/L

3 mol Ca(OH)2

1 mol Al2(SO4)3 x

# L Ca(OH)2=0.0250

L Al2(SO4)3

= 0.375 L Ca(OH)2

0.125 mol Al2(SO4)3 L Al2(SO4)3

x

2. Al2(SO4)3(aq) + 3Ca(OH)2(aq) 2Al(OH)3(s) + 3CaSO4(s)

L Ca(OH)2

0.0250 mol Ca(OH)2

x

Answers

3 mol Na2CO3

2 mol FeCl3

x

# L Na2CO3=

0.0750 L FeCl3

= 0.0900 L Na2CO3 = 90.0 mL Na2CO3

0.200 mol FeCl3

L FeCl3

x

3. 2FeCl3(aq) + 3Na2CO3(aq) Fe2(CO3)3(s) + 6NaCl(aq)

L Na2CO3

0.250 mol Na2CO3

x

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 M NaOH?

H2SO4(aq) + 2NaOH(aq) 2H2O + Na2SO4(aq)

2. How many moles of Fe(OH)3 are produced when 85.0 L of iron(III) sulfate at a concentration of 0.600 M reacts with excess NaOH?

Fe2(SO4)3(aq) + 6NaOH(aq) 2Fe(OH)3(s) + 3Na2SO4(aq)

3. What mass of precipitate will be produced from the reaction of 50.0 mL of 2.50 M sodium hydroxide with an excess of zinc chloride solution.

2NaOH(aq) + ZnCl2(aq) Zn(OH)2(s) + 2NaCl(aq)

Assignment

4. a) What volume of 0.20 M AgNO3 will be needed to react completely with 25.0 mL of 0.50 M potassium phosphate?

3AgNO3(aq) + K3PO4(aq) Ag3PO4(s) + 3KNO3(aq)

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 M aluminum nitrate solution is mixed with 0.240 L of 1.50 M sodium hydroxide solution?

Al(NO3)3(aq) + 3NaOH(aq) Al(OH)3(s) + 3NaNO3(aq)

Assignment

Answers

2 mol Fe(OH)3

1 mol Fe2(SO4)3 x

# mol Fe(OH)3=

85 L Fe2(SO4)3 0.600 mol Fe2(SO4)3

L Fe2(SO4)3 x

1. H2SO4(aq) + 2NaOH(aq) 2H2O + Na2SO4(aq)

= 102 mol

1 mol H2SO4

2 mol NaOHx

# L H2SO4=0.075 L

NaOH

= 0.009375 L = 9.4 mL

0.50 mol NaOH L NaOH

x

2. Fe2(SO4)3(aq) + 6NaOH(aq) 2Fe(OH)3(s) + 3Na2SO4(aq)

L H2SO4

2.0 mol H2SO4

x

3. 2NaOH(aq) + ZnCl2(aq) Zn(OH)2(s) + 2NaCl(aq)

1 mol Zn(OH)2

2 mol NaOHx

# g Zn(OH)2=

0.0500

L NaOH

= 6.21 g2.50 mol NaOH

L NaOHx

4a. 3AgNO3(aq) + K3PO4(aq) Ag3PO4(s) + 3KNO3(aq)

99.40 g Zn(OH)2

1 mol Zn(OH)2

x

3 mol AgNO3

1 mol K3PO4

x

# L AgNO3 =

0.025

L K3PO4

= 0.1875 L = 0.19 L0.50 mol K3PO4

L K3PO4

x L AgNO3

0.20 mol AgNO3

x

1 mol Ag3PO4

1 mol K3PO4

x

# g Ag3PO4=

0.025

L K3PO4

= 5.2 g0.50 mol K3PO4

L K3PO4

x 418.58 g Ag3PO4

1 mol Ag3PO4

x

4b. 3AgNO3(aq) + K3PO4(aq) Ag3PO4(s) + 3KNO3(aq)

1 mol Al(OH)3

1 mol Al(NO3)3

x

# g Al(OH)3=0.550 L Al(NO3)3

21.4 g Al(OH)3=

77.98 g Al(OH)3

1 mol Al(OH)3

x 0.500 mol Al(NO3)3

L Al(NO3)3

x

5. Al(NO3)3(aq) + 3NaOH(aq) Al(OH)3(s) + 3NaNO3(aq)

1 mol Al(OH)3

3 mol NaOHx

# g Al(OH)3=0.240

L NaOH

9.36 g Al(OH)3=

77.98 g Al(OH)3

1 mol Al(OH)3

x 1.50 mol NaOHL NaOH

x

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