As you come in,
 The



Remote control
Pick up packet.
Paper, pencil, calculator, periodic table for
notes
 The


Plan:
Learn about stoichiometry
 The

Materials:
Assessments:
Stoichiometry Quiz on Thursday
Limiting Reactants Quiz on Friday
Halloween Pandora Station – Project Eliminate Challenge
Unit 7
Stoichiometry
Chapter 9
Info in a Chemical Equation
2K + CuSO4  Cu + K2SO4
A
balanced chemical equation shows the
correct ratios required for a chemical
reaction to occur.
 Knowing the ratio allows us to provide the
appropriate amount of reactants AND
predict the amount of products.
Mole to Mole Relationships
 Since
chemical equations give
appropriate relationships of moles
of each compound, mole ratios
can be written.
 Mole ratio: uses the coefficients in
the balanced equation to show
how two compounds are related in
a reaction
Example 9.2
Goal: Write a mole ratio & solve the problem.
What
number of moles of O2
will be produced by the
decomposition of 5.8 mol of
water?
 First,
the balanced chemical
equation must be written for the
situation described in the prompt.
 2H2O  2H2 + O2
Example 9.2 (continued)
What
number of moles of O2 will
be produced by the
decomposition of 5.8 mol of
water?
 Next,
2H2O  2H2 + O2
the reaction shows that oxygen and
water have a quantitative relationship: a
mole ratio.
1 mole O2 : 2 moles H2O
 Where did the numbers (1 & 2) come
from?
 Can you write other mole ratios from this
equation?
Example 9.2 (continued)
What
number of moles of O2 will
be produced by the
decomposition of 5.8 mol of
water?
 Finally,
2H2O  2H2 + O2
we can solve this problem using the
mole ratio as a conversion factor in our
dimensional analysis.
 Start
with the given.
 Set up the chart to cancel units AND
compounds.
 Multiply or divide as usual.
Example 9.3
Goal: Solve the problem with a partner.
Calculate
the number of moles of
oxygen required to react exactly
with 4.30 mol of propane, C3H8, in
the reaction described by the
following:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
Calculate the number of moles of
oxygen required to react exactly with
4.30 mol of propane, C3H8, in the
reaction described by the following:
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
A.) 0.860 moles of O2
B.) 7.17 moles of O2
C.) 21.5 moles of O2
D.) none of these
Calculate the number of moles of
oxygen required to react exactly with
4.30 mol of propane, C3H8, in the
reaction described by the following:
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
A.) 0.860 moles of O2
B.) 7.17 moles of O2
C.) 21.5 moles of O2
D.) none of these
Example 9.4
Is another example needed?
 Ammonia
is used in huge quantities as a
fertilizer. It is manufactured by combining
nitrogen and hydrogen according to the
following equation:
N2(g) + 3H2(g) = 2NH3(g)
Calculate the number of moles of NH3 that
can be made from 1.30 mol H2(g) reacting
with excess N2(g).
Suggested Homework
 Page
287 4 and 5
Mega Mole Map
 Mole
to mole relationship
 Mole Ratio = conversion factor (bridge)
 In
our examples so far, we’ve been given
moles of one compound and asked to
convert to moles of another compound.
 What if I told you that I won’t always start
out with moles or ask for moles? Starting
or ending with mass, liters, or particles is
very common.
 Work with your partner to expand on our
current Mole Concept Map.
Mega Mole Map

Mole to mole relationship
 Mole Ratio = conversion factor (bridge)

Mole to mass relationship
 Molar Mass = conversion factor (bridge)

Mole to particle relationship
 Avogadro’s # = conversion factor (bridge)

Mole to volume relationship
 Molar volume = conversion factor (bridge)
Example 9.5
Goal: Use new mole concepts to solve the
problem.
 Consider
the reaction of powdered
aluminum metal and finely ground iodine
to produce aluminum iodide. The
balanced equation for this vigorous
chemical reaction is:
2Al(s) + 3I2(s) = 2AlI3(s)
Calculate the mass of I2(s) needed to just react
with 35.0 g of Al(s).
 First,
write the given and set up your chart.
Example 9.5 (continued)
 Consider
the reaction of powdered
aluminum metal and finely ground iodine to
produce aluminum iodide. The balanced
equation for this vigorous chemical reaction
is:
2Al(s) + 3I2(s) = 2AlI3(s)
Calculate the mass of I2(s) needed to just react with
35.0 g of Al(s).
 Next,
consult your map to determine a problemsolving path. How many steps will this problem
require?
 3 steps
Example 9.5 (continued)
 Consider
the reaction of powdered
aluminum metal and finely ground iodine to
produce aluminum iodide. The balanced
equation for this vigorous chemical reaction
is:
2Al(s) + 3I2(s) = 2AlI3(s)
Calculate the mass of I2(s) needed to just react with
35.0 g of Al(s).
 Finally,
plug in your conversion factors. Multiply
and divide as usual to give your final answer.
 493.89 g I2
Section 9.2 Review Questions
Goal: Solve the problem with a
partner.
2.
Solutions of sodium hydroxide cannot
be kept for very long because they
absorb carbon dioxide from the air,
forming sodium carbonate. The
unbalanced equation is:
NaOH(aq) + CO2(g)  Na2CO3(aq) + H2O(l)
Calculate the number of grams of
carbon dioxide that can be absorbed
by complete reaction with a solution
that contains 5.00 g of NaOH.
NaOH(aq) + CO2(g)  Na2CO3(aq) + H2O(l)
Calculate the number of grams of carbon
dioxide that can be absorbed by complete
reaction with a solution that contains 5.00 g
of NaOH.
A.) 2.75 g CO2
B.) 5.50 g CO2
C.) 9.09 g CO2
D.) none of these
NaOH(aq) + CO2(g)  Na2CO3(aq) + H2O(l)
Calculate the number of grams of
carbon dioxide that can be absorbed
by complete reaction with a solution
that contains 5.00 g of NaOH.
A.) 2.75 g CO2
B.) 5.50 g CO2
C.) 9.09 g CO2
D.) none of these
Suggested Homework
 11-2
Practice Problems – (1-11…skip 9)
NOTE: We’ll use iRespond TODAY to check
your answers, and we’ll make colored
pencil corrections on your work.
Limiting Reactants
 As
you know, a balanced chemical
equation gives the perfect ratio of
reactants needed to perform the
reaction.
 In reality, we rarely have the “perfect”
ratio.
 With an imperfect ratio, one reactant will
run out before the other. The reactant
running out will stop the reaction...limit
the products.
 The reactant that runs out = Limiting
reactant
 The reactant that remains = Excess
reactant
Limiting Reactant Calculations
 Using
stoichiometry, we can calculate the
following predictions:




Identify the limiting reactant
Identify the excess reactant
Predict the amount of product to form
Predict the amount of excess reactant
remaining
Practice Problem 9.5
 Calculate
the mass of AlI3(s) formed by
the reaction of 35.0 g Al(s) with 495 g I2.
Example 9.7
 Suppose
that 25.0 kg of nitrogen
gas and 5.00 kg of hydrogen gas
are mixed and reacted to form
ammonia. Calculate the mass of
ammonia produced when this
reaction is run to completion.
 Two
“givens” means two calculations
 The unknown is the same in the two
calculations
 The smallest answer is the best prediction as it
tells when the limiting reactant will run out.
Follow-up Questions
 Identify
the limiting reactant.
 Identify the excess reactant.
 How much H2 is used? Left?
 How much N2 is used? Left?
Limiting Reactant Practice
1.The reaction between solid white phosphorus and
oxygen produces solid tetraphosphorus decaoxide
(P4O10). This compound is often called phosphorus
pentoxide because its empirical formula is P2O5.
P4 + 5O2  P4O10
a.
b.
c.
Determine the mass of tetraphosphorus decaoxide
formed if 25.0 g of phosphorus (P4) and 50.0 g of
oxygen gas are combined.
Identify the limiting reactant.
How much of the excess reactant remains after the
reaction stops?
FYI
 "Stoichiometry"
is derived from the Greek
words στοιχεῖον (stoicheion, meaning
element]) and μέτρον (metron, meaning
measure.)
Section 9.2
Review Questions
4.
a.
b.
You react 10.0 g of nitrogen gas
with hydrogen gas according to
the following reaction.
N2(g) + 3H2(g) = 2NH3(g)
What mass of hydrogen is required
to completely react with 10.0 g
sample of nitrogen gas?
What mass of ammonia is
produced from 10.0 g of nitrogen
gas and sufficient hydrogen gas?
Practice Problem 9.8
 Lithium
nitride, an ionic compound
containing the Li+ and N3- ions, is prepared
by the reaction of lithium metal and
nitrogen gas. Calculate the mass of
lithium nitride formed from 56.0 g of
nitrogen gas and 56.0 g of lithium.
Follow-up Questions
 Identify
the limiting reactant.
 Identify the excess reactant.
 How much of each reactant is used?
Left?
Percent Yield
 Yield
means product.
 Calculating % yield is calculating what %
of the product your experiment actually
produced.
 You are comparing your prediction to
your action.
% Yield
 (Actual
yield / theoretical yield) 100
 Actual yield = either given in the problem
or carried out in the lab
 Theoretical yield = ALWAYS calculated by
stoichiometry
Stoichiometry Test Tips
1.
2.
3.
Know how to write and balance a
combustion reaction. (See #1 on
your Stoichiometry Quiz.)
Remember, “a liter” or “a single
gram” means 1 L or 1 gram. The
given can be written this way.
“How many grams of product are
produced?” - refers to theoretical
yield (smallest answer) in a limiting
reactant problem
4. Metal products are solids.
5. STP = Standard temperature and
pressure (doesn’t affect the
calculation)
6. % yield = (ACTUAL/THEORETICAL) 100
7. Excess reactant - Use dimensional
analysis starting with given
amount of LR to calculate the
amount USED; Subtract from
ORIGINAL amount to calculate
excess LEFT
Look closely at the
wording…
 Labeling

the equation will help!
If you are given info about BOTH reactants, then
you have a limiting reactant problem.
 Example:
What mass of NaCl will be produced by the
reaction of 58.7g of NaI with 29.4g of Cl2 gas if the
products are NaCl and I2?

If you are given info about ONLY ONE substance,
then you have a simple stoichiometry problem.
More wording advice…
 LABELS!

If you are given info about BOTH reactants
AND a product, then you’ve probably got
a percent yield problem to solve. The
product info is the actual yield.
 Determine
the percent yield for a reaction
between 6.92g K and 4.28g of O2 if 7.36g of
K2O is produced.