Stoichiometry

Ch.10

(10-1) Stoichiometry

• Mass & amt relationships b/w reactants & products– Conversions b/w grams & moles

• Always begin w/ a balanced eq.!

Mass to Mol Reminder

Convert 3.5 g of NaOH to mols NaOH1. List known

3.5 g NaOH

2. Calculate molar mass 22.99 g/mol Na + 16 g/mol O + 1.01 g/mol H = 40 g/mol

3. Set up problem & solve 3.5 g NaOH x 1 mol NaOH = 0.09 mol NaOH

40 g NaOH

Mol to Mass Reminder

How many grams are in 6.4 mols O2?1. List known

6.4 mols O2

2. Calculate molar mass(2) (16 g/mol O) = 32 g/mol O2

3. Set up problem & solve 6.4 mols O2x 32 g O2 = 204.8 g O2

1 mol O2

Mol to Mol Reminder

If 2.00 mol N2 is reacting w/ a sufficient amt of H2, how many mols of NH3 will be produced?

N2 + H2 NH3

1. Balance the chemical eq.N2 + 3H2 2 NH3

Mol to Mol Reminder

2. Find the mole ratio1 mol N2 : 2 mol NH3

3. Set up problem (begin w/ known) & solve

2.00 mol N2 x 2 mol NH3 = 4 mol NH3

1 mol N2

Conversions

g of A mol of A mol of B g of B

Mole ratio (coeff.from

bal.eq.)

Molar Mass(from PT)

Molar Mass(from PT)

Mass to Mass Practice

How many grams of H2O are produced when 13 g O2 combine w/ sufficient H2?

H2 + O2 H2O

1. Balance the chemical eq.

2 H2 + O2 2 H2O

Mass to Mass Practice

2. Calculate molar mass of known & unknown

O2 = 32 g/mol

H2O = 18.02 g/mol

3. Find mole ratio1 mol O2 : 2 mol H2O

Mass to Mass Practice

4. Set up problem (begin w/ known) & solve

13 g O2 x 1 mol O2 x 2 mol H2O x 18.02 g H2O

32 g O2 1 mol O2 1 mol H2O

= 14.63 g H2O

Using Density

• Density (D) = mass (m) / volume (V)

• Units: g/mL

• Convert from g mL or mL g

Density Practice

Calculate the mass of LiOH used to obtain 1500 mL of water. (Hint: DH2O = 1.00 g/mL)

CO2 + 2LiOH Li2CO3 + H2O

1. Start w/ known & use density to convert into grams

1500 mL H2O x 1.00 g H2O

1 mL H2O

Density Practice

2. Proceed w/ mass to mass calc.

1500 mL H2O x 1.00 g H2O x 1 mol H2O x

1 mL H2O 18.02 g H2O

2 mol LiOH x 23.95 g LiOH = 3989 g LiOH

1 mol H2O 1 mol LiOH

(10-2) Excess Reactant

• Extra left over after rxn

• Limiting reactant: completely consumed – Limits amt of other reactants used– Determines max amt of product

The Cheese Sandwich AnalogyThe Cheese Sandwich Analogy

Limiting Reactant Practice

CO combines w/ H2 to produce CH3OH. If you had 152.5 g CO & 24.5 g H2 what mass of CH3OH could be produced?

1. Write a bal. eq.CO + 2 H2 CH3OH

Limiting Reactant Practice

2. Convert reactants to mols present

152.5 g CO x 1 mol CO = 5.444 mol CO present

28.01 g CO

24.50 g H2 x 1 mol H2 = 12.1 mol H2 present

2.02 g H2

Limiting Reactant Practice

3. Using the reactants mol ratio, find how many mols needed

12.1 mol H2 x 1 mol CO = 6.06 mol CO needed

2 mol H2

• CO present is not enough to react w/ all the H2, so CO is limiting

Limiting Reactant Practice

4. Use limiting reactant to set up stoich.5.444 mol CO x 1 mol CH3OH x 32.05 g CH3OH

1 mol CO 1 mol CH3OH

= 174.5 g CH3OH

Theoretical Yield

• Calculated max amt of product possible– Stoich. w/ limiting reactant– What should happen

• Actual yield: measured amt of product experimentally produced – What does happen

Percentage Yield

• How efficient a rxn is– How close actual is to theoretical– Should be 100%

• % yield = actual x 100

theoretical

% Yield Practice

When 0.835 mol LiOH is reacted w/ excess KCl, the actual yield of LiCl is 16 g. What is the % yield?

LiOH + KCl LiCl + KOH

% Yield Practice

1. Calculate the theor. yield0.835 mol LiOH x 1 mol LiCl x 42.44 g LiCl = 35.4 g LiCl

1 mol LiOH 1 mol LiCl

2. Calculate the % yield% yield = act. X 100 = 16 g LiCl x 100 = 45 %

theor. 35.4 g LiCl