Solution types of stoichiometry problems are no harder than any other stoichiometry problem. You must use the concentration given (molarity) to convert to moles, then use the molar ratio to convert to moles of the final product, then use concentration of the final product to convert to the final answer.

Concentration (given) Moles of Given Moles of Unknown Concentration (unknown)

Moles/L Moles/L

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

Calculate the volume of a 2.00 mol/L silver nitrate solution that is needed for 12.0 g of copper metal to react according to the following equation?

2 mol AgNO3

1 mol Cu x

# mol AgNO3 =

12.0 g Cu 0.38 mol AgNO3= 1 mol Cu 63.5 g Cu

x

Cu(s) + 2AgNO3(aq) Cu(NO3)2 + 2Ag(s)

Calculate the moles of AgNO3, then volume of AgNO3

L = 0.38 mol AgNO3 / 2.00 mol / L = 0.19L = 190mL

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?

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

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.

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)20.0250 mol

Ca(OH)2

x

A chemistry teacher wants 75.0 mL of 0.200 mol/L iron(Ill) chloride solution to react completely with a 0.250 mol/L 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

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 silver phosphate is produced from the above reaction?

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 NaOHx

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 L

0.50 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)