1 Pb(NO3)2(aq) + KI (aq) PbI2(s) + KNO3(aq)__ __ __ __2 1 2

Molarity and Stoichiometry

M M

V V

P P

mol mol

M L M L

M =mol

L

mol = M L

What volume of 4.0 M KI solution is required to yield 89 g PbI2?

Step 1) Identify the species present in the combined solution, and determine what reaction occurs.

Step 2) Write the balanced net ionic equation for the reaction.

Step 3) Calculate the moles of reactants.

Step 4) Determine which reactant is limiting.

Step 5) Calculate the moles of product or products, as required.

Step 6) Convert to grams or other units, as required.

Stoichiometry steps for reactions in solution

Stoichiometry for Reactions in Solution

461 g PbI2

Strategy:

1 Pb(NO3)2(aq) + 2 KI (aq) 1 PbI2(s) + 2 KNO3(aq)

X mol KI = 89 g PbI2

1 mol PbI2

1 mol PbI2

2 mol KI= 0.39 mol KI

(1) Find mol KI needed to yield 89 g PbI2.(2) Based on (1), find volume of 4.0 M KI solution.

What volume of 4.0 M KI solution is required to yield 89 g PbI2?

89 g? L 4.0 M

M = molL

L = molM

= 0.39 mol KI

4.0 M KI= 0.098 L of 4.0 M KI

= 0.173 mol CuSO4

How many mL of a 0.500 M CuSO4 solution will react w/excess Al to produce 11.0 g Cu?

__CuSO4(aq) + __Al (s)

Al3+ SO42–

CuSO4(aq) + Al (s) Cu(s) + Al2(SO4)3(aq)3 2 3 1x mol 11 g

__Cu(s) + __Al2(SO4)3(aq)

X mol CuSO4 = 11 g Cu1 mol Cu

63.5 g Cu

3 mol CuSO4

3 mol Cu

M = molL

L = molM

0.173 mol CuSO4

0.500 M CuSO4

= 0.346 L

0.346 L 1000 mL

1 L= 346 mL

63.55g Cu

1 molCu

Stoichiometry Problems• How many grams of Cu are required to react

with 1.5 L of 0.10M AgNO3?

1.5L

.10 molAgNO3

1 L= 4.8 g

Cu

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

1 molCu

2 molAgNO3

? g 1.5L0.10M

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants• 79.1 g of zinc react with 0.90 L of 2.5M HCl.

Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP?

Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L

2.5M

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants

79.1g Zn

1 molZn

65.39g Zn

= 27.1 L H2

1 molH2

1 molZn

22.4 LH2

1 molH2

Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L

2.5M

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants

22.4L H2

1 molH2

0.90L

2.5 molHCl

1 L= 25 L

H2

1 molH2

2 molHCl

Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L

2.5M

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants

Zn: 27.1 L H2 HCl: 25 L H2

Limiting reactant: HCl

Excess reactant: Zn

Product Formed: 25 L H2

left over zincCourtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem