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