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Stoichiometry:
Calculations with Chemical
Equations

Objectives
 Use chemical equations to predict amount of
product from given reactants
 Determine percentage yield
 Determine limiting reactant

Working with equations:
STOICHIOMETRY
 Predict how much product is obtained from
given amount of reactant
 Predict how much reactant is needed to give
required amount of product
 Predict how much of one reactant is
required to give optimum result with given
amount of another reactant

Relating moles, masses and
molecules
 Note:
 Conservation of mass (4 g + 32 g = 36 g)
 But not necessarily conservation of moles (2
moles + 1 mole = 2 moles)

Stoichiometry with equations:
The roadmap




Equations are in moles, but we measure in grams
Three conversions required:
A is given substance; B is target substance
1
...
Use coefficients in equation to get moles B from moles A
3
...
Convert moles A → moles B:
Mole:mole ratio (target/given):
moles B
moles A



b

mol B / mol A

a

2
...
Mass A → moles A using molar mass A
2
...
Moles B → mass B using molar mass B

Mass A

Mass/molar mass

Moles A

Mole:mole ratio

Moles B

Moles x molar
mass

Mass B

 Molar mass Cl2 = 35
...
0 g/mol
 Molar mass NaOH = 23
...
00 + 1
...
01 g/mol

Summary of stoichiometry problems


Maximum of three conversions required
1
...
Use coefficients in equation to get moles B from
moles A
3
...
Molar mass of given substance (maybe)
2
...
Balanced chemical equation (always)

Work this example
 CH4 + 2O2 = CO2 + 2H2O
 What mass of CO2 is produced by the
complete combustion of 16 g of CH4
 Atomic weight H = 1, C = 12, O = 16

44 g
 Do stoichiometry exercises

Reaction Yield
 Actual yield from chemical reaction is
normally less than predicted by
stoichiometry
...
8 g after reaction of 26
...

 What is theoretical yield? Use stoichiometry to get mass of
product:
convert mass (26
...
4 g
 Percent yield = 32
...
4 x 100 %

 Do percent yield exercises
Mass A

Mass/molar mass

Moles A

Mole:mole ratio

Moles B

Moles x molar
mass

Mass B

Percent yield practice

Limiting Reactant
 Exact quantities of reactants dictated by the
reaction stoichiometry are not the norm
 Usually one reactant is reacted with an
excess of the other(s)
 Burning natural gas in furnace

 This reactant is the limiting reactant –
amount of products limited by this reactant

Limiting reactant at molecular level
 In the reaction to produce ethylene glycol:
 1 mole of ethylene oxide + 1 mole of water → 1 mole glycol

 Here, 2 moles of water remain after conversion of all
ethylene oxide into glycol
 Ethylene oxide is limiting

 More product would be obtained by increasing the ethylene
oxide until the water became limiting

Determination of limiting reactant
 Two methods:
 Brute Force
 Calculate quantity of product from each reactant
in turn

 Elegant
 Compare reaction stoichiometry with actual
reactant mole ratios

The Brute Force method
 30 g CH4 and 30 g O2 are reacted

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