1. Chem Ch 9 Notes
• 9.1 Stoichiometry- calculation of quantities
in chemical reactions
• From the coefficients in a BALANCED
equation you can get:
• N2+3H2 2NH3
• 1. # particles (molecules, formula units,
atoms)
• 1 molecule N2: 3 molecules H2 : 2
molecules NH3
• 2. # moles
• 1 mole N2: 3 mole H2 : 2 mole NH3

2. • 3. Mass- Obeys Law of Conservation of
Mass
• 1 mol N2 = 28.0 g
• +3 mol H2 = 6.0 g
• 2 mol NH3 = 34.0 g
• 4. Volume- GASES at STP 1 mol= 22.4L
• 3 mol H2 = 3(22.4)=67.2L
• 1 mol N2 = 1(22.4) = 22.4 L
• 2 mol NH3 = 2(22.4) = 44.8 L
• Notice that the number of atoms & mass
are always conserved, but the volume &
molecules may not be!!

3. • Mole-Mole calculations
• N2+3H2  2NH3
• 1mol N2 1moleN2 2 mol NH3
3 mol H2 2mol NH3 3 mol H2
These are your mole conversion factors
How many moles of NH3 are in 0.60 mol
N2?

4. • Mass:Mass
• 1. Convert mass to moles (1mol/ g)
• 2. Convert moles of what you have to moles
of what you want
• 3. Convert mol to mass ( g/1mol)
• How many g NH3 are produced when 5.40 g
of H2 react w/ excess N2?
• For all problems
• Convert to mol, use mol:mol, convert to
wanted units

5. • How many moles O2 are produced when 29.2
g H2O decomposed into O2 & H2?
• Assuming STP, how many L O2 are needed
to produce 19.8 L SO3?
• 2SO2(g) + O2(g)  2 SO3(g)

6. • Use ratios from coefficients b/c the
volumes relate from 22.4L=1mol
• 2NO(g) + O2 (g)  2 NO2(g)
• How many mL NO2 are produced when
3.4mL O2 react with excess NO at STP?
• 1mL O2=2 mL NO2=2 mL NO

7. • Limiting reactant & % Yield
• Limiting Reagent- determines amt. Of
product formed, runs out 1st
• Steps: (if given 2 reactants)
• 1. Solve for same product
• 2. One w/least amount of product is
limiting
• 3. Other is excess (left over)
• 4. Limiting determines the amount of
product

8. • 2 Na(s) + Cl2 (g)  2NaCl (s)
• Suppose 8.70 mol Na react w/3.20 mol Cl2.
• What is limiting reactant?
• How many moles of NaCl are produced?
• How many g NaCl are produced?
• How much of the excess reactant is
left?

9. • 2Cu (s) + S (s)  Cu2S(s)
• What is the limiting reactant when 80.0
g Cu reacts with 25.0 g S?
• How many g Cu2S can be formed?
• How much of the excess reactant is
left?

10. • % yield- measures the efficiency of
reaction
• Normally can’t be larger than 100%
• Most are less than 100 % b/c:
• Reactions don’t go to completion
• Impure reactants
• Competing side reaction
• Loss of product during transfer
or filtration
• Human error (bad measurement,
miscalculation)

11. • Actual yield- amt. of product formed in lab
• Theoretical yield- amt. product that could
be formed from calculations
• % Yield: actual yield X 100
theoretical yield
• CaCO3 (s)  CaO (s) + CO2 (g)
• What is the theoretical yield of CaO if
24.8 g CaCO3 is heated?
• What is the % yield if 13.1 g CaO is
produced?