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Unit 12: Stoichiometry
Chapter 12
Chemistry 1K
Cypress Creek High School
Reminders…
• The mole is a counting unit
– 1 mole = 6.02 x 1023 particles (atoms,
molecules, formula units)
• Molar mass is the mass of 1 mole
– Measured in grams
– Represented by the atomic mass shown
on the periodic table
• Example: 1 mole of water (H2O) is
about 18 grams. It contains 2 moles
of hydrogen atoms and 1 mole of
oxygen atoms.
Mole Map
What is Stoichiometry?
• Using the methods of stoichiometry, we can
measure the amounts of substances involved in
chemical reactions and relate them to one another.
– Pronounced stoy-kee-AW-muh-tree
– From the Greek words stoikheion meaning
element and metria meaning measure
Stoichiometric Calculations
• There are several basic stoichiometric
calculations:
– mole-to-energy conversions
– mole-to-mole conversions
– mole-to-mass / mass-to-mole conversions
– mass-to-mass conversions
– mole-to-volume conversions
• All stoichiometric calculations are based on a
balanced equation and mole ratios.
Mole Ratios
• Mole ratios indicate the molar relationship between
two chemicals in an equation
– Use the coefficients (not subscripts) to make ratios
– Reduce to lowest whole-number ratio
• Example: Water Formation
– 2H2 + O2  2H2O
• What is the ratio of hydrogen to oxygen? 2:1
• What is the ratio of oxygen to water? 1:2
• What is the ratio of water to hydrogen? 1:1
Mole-to-Mole Conversions
• The following reaction shows table salt production.
How many moles of sodium chloride are produced
from 0.02 moles of chlorine?
Solve through ratios:
X mol NaCl = 2 mol NaCl
0.02 mol Cl2 1 mol Cl2
(cross multiply and solve) =
0.04 mol NaCl
Mole-to-Energy Conversions
• In the decomposition of baking soda, NaHCO3(s)
+ 1,800 kJCO2(g) + NaOH(s), how much
energy is required if there is 3 moles of NaOH
produced?
Solve through ratios:
X kJ = 3 mol NaOH
1800kJ 1 mol NaOH
(cross multiply and solve) =
5400kJ
Mole-to-Volume Conversions
• Given 1 mol of O2, what volume of NO is needed
in the reaction: NO + O2  NO3?
Solve through ratios:
X L NO = 1mol O2
1(22.4)L H2
1 mol O2
(cross multiply and solve) =
22.4L NO
Volume-to-Volume Conversions
• N2 + 3H2  2NH3, what volume of hydrogen is
necessary to react with 5L of nitrogen to produce
ammonia?
Solve through ratios:
X L H2 =
5L N2
3(22.4)L H2 1(22.4)L N2
(cross multiply and solve) =
15L
Mole-to-Mass Conversions
• The following reaction shows photosynthesis. How many
grams of glucose are produced when 24 moles of carbon
dioxide reacts with water?
Solve through ratios:
X g C6H12O6 =
24 mol CO2
180 g C6H12O6 (cross multiply and solve) =
720 g C6H12O6
6 mol CO2
Mass-to-Mole Conversions
• The following reaction shows photosynthesis. How many
moles of glucose are produced when 132 grams of carbon
dioxide reacts with water?
Solve through ratios:
X mol C6H12O6 = 1 mol C6H12O6
132 g CO2
6(44 g CO2)
(cross multiply and solve) =
0.5 mol C6H12O6
Mass-to-Mass Conversions
• The following reaction shows the production of
ammonia. How many grams of nitrogen are
required to produce 85 grams of ammonia?
Solve through ratios:
X g N2 = 1(28 g N2)
85 g NH3 2(17 g NH3)
(cross multiply and solve) =
70 g N2
Limiting Reactant
• In the real world, reactants are not present in the
exact mole ratio described by the balanced
equation.
• This means that one of the reactants will be used
up before the other one.
– The limiting reactant is used up first and
restricts (stops) the reaction
– The excess reactant(s) remain after the reaction
stops
• Example: If you placed a single drop of water on an
Alka-Seltzer tablet…
– What would be the limiting reactant?
– Excess reactant?
– How could you switch the roles of these two
reactants?
Limiting Reactant
• Experiment: Equal amounts of copper (II) chloride are
combined with varying amounts of aluminum to form
copper and aluminum chloride. Which pairing will react
completely?
• 2 Al(s) + 3 CuCl2(aq)  2 AlCl3(aq) + 3 Cu(s)
A
B
C
1:1 mass
1:1 mole
2:3 mass
D
Al:CuCl2
ratio
2:3 mole
Limiting Reactant
• Results: In reactions A, B, & C, aluminum is still visible the excess reactant. Copper (II) chloride is the limiting
reactant. Reaction D, which follows the correct ratio in
the balanced equation, reacted completely.
• 2 Al(s) + 3 CuCl2(aq)  2 AlCl3(aq) + 3 Cu(s)
A
B
C
D
Al:CuCl2
ratio
1:1 mass
1:1 mole 2:3 mass 2:3 mole
Limiting Reactant Practice
• Diagram A represents reactant particles, and
Diagram B represents product particles in a
fictitious chemical reaction.
Write the
balanced
Diagram A
Diagram B
equation for
this reaction.
A
Z
C
A
Z
C
Z
A
A
A
A
A
A
A
Z
C
Z
A Z
A
A
Z
C
Z
A
A
What is the
limiting
reactant?
Excess?
Limiting Reactant Practice
• The three particle diagrams
to the right represent the
burning of methane. In
which diagram is the limiting
reactant oxygen gas?
CH4 + 2O2  CO2 + 2 H2O
A
B
C
Limiting Reactant Calculations
Method 1
• In a reaction, 80 grams of sodium hydroxide is combined with 60
grams of sulfuric acid. What is the limiting reactant?
• Convert one of the masses of a given reactants into the mass of
the other:
Xg NaOH
= 60g H2SO4
= 48.98g NaOH
2(40g NaOH) 1(98g H2SO4)
• Compare the mass calculated to the mass given in the problem
– 48.98g of NaOH is needed to react with 60 g of H2SO4
– You are given 80g of NaOH, therefore there is more NaOH than
required to react with 60g of H2SO4
– Sulfuric acid is the limiting reactant
Limiting Reactant Calculations
Method 2
• In a reaction, 80 grams of sodium hydroxide is combined with 60
grams of sulfuric acid. What is the limiting reactant?
• Calculate the moles of one of the product from each reactant
(use H2O if present):
X mol H2O = 80 g NaOH
2 mol H2O 2(40g NaOH)
X = 2 mol H2O
x mol H2O = 60 g H2SO4
2 mol H2O 1(98g H2SO4)
x = 1.22 mol H2O
• Compare the 2 molar amounts
– Sulfuric acid produces less water
– Sulfuric acid is the limiting reactant
Percent Yield
• Percent yield compares the amount of product
collected in an experiment (actual) to the
amount anticipated according to calculations
(theoretical).
• Actual mass will always be less than theoretical
mass due to human error
actual mass of product
x 100
% yield =
theoretical mass of product
Percent Yield Calculations
• Methanol, (CH3OH), was used as one of several embalming
fluids in ancient Egypt. In an experiment, you used 25 grams
of CO2 with excess hydrogen gas to produce 15 grams of
methanol. What is the percent yield?
• Calculate theoretical yield
X g CH3OH = 25 g CO2 = 18.18 g
1( 32g CH3OH) 1(44 g CO2)
• Calculate percent yield
15 g CH3OH x 100 = 82.5% yield
18.18 g CH3OH