STOICHIOMETRY TUTORIAL
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(1-2-3) General Approach For Problem Solving:
1. Clearly identify the Goal or Goals and the UNITS involved. (What
are you trying to do?)
2. Determine what is given and the UNITS.
3. Use conversion factors (which are really ratios) and their UNITS to
CONVERT what is given into what is desired.
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Sample problem for general problem solving.
Sam has entered into a 10 mile marathon. Use ALL of the following
conversions (ratios) to determine how many inches there are in the race.
5280 ft = 1 mile; 12 inches = 1 ft
1. What is the goal and what units are needed?
Goal = ______ inches
2. What is given and its units?
10 miles
3. Convert using factors (ratios).
10 miles
Given
Units match
=
633600 inches
Goal
Convert
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Sample problem #2 on problem solving.
A car is traveling at a speed of 45 miles per hr (45 miles/hr). Determine
its speed in kilometers per second using the following conversion
factors (ratios). 1 mile = 5280 ft; 1 ft = 12 in; 1 inch = 2.54 cm; km = 1
x 103; cm = 1 x 10-2m; 1 hr =60 min; 1 min = 60 s
Goal
Given
= 0.020
This is
the
same as
putting
k over k
c cancels c
m remains
km
s
Units Match!
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Converting grams to moles.
Determine how many moles there are in 5.17 grams of Fe(C5H5)2.
Given
Goal
units match
5.17 g Fe(C5H5)2
= 0.0278
Use the molar mass
to convert grams to
moles.
moles Fe(C5H5)2
Fe(C5H5)2
2 x 5 x 1.001 = 10.01
2 x 5 x 12.011 = 120.11
1 x 55.85 = 55.85
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Stoichiometry (more working with ratios)
Ratios are found within a chemical equation.
2HCl + 1Ba(OH)2 → 2H2O + 1 BaCl2
coefficients give MOLAR RATIOS
2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O
and 1 mole of BaCl2
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Mole – Mole Conversions
When N2O5 is heated, it decomposes:
2N2O5(g) → 4NO2(g) + O2(g)
a. How many moles of NO2 can be produced from 4.3 moles of N2O5?
2N2O5(g) → 4NO2(g) + O2(g)
4.3 mol
? mol Units match
4.3 mol N2O5
= 8.6
moles NO2
b. How many moles of O2 can be produced from 4.3 moles of N2O5?
2N2O5(g) → 4NO2(g) + O2(g)
4.3 mol
? mol
4.3 mol N2O5
= 2.2
mole O2
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gram ↔ mole and
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gram ↔ gram conversions
When N2O5 is heated, it decomposes:
2N2O5(g) → 4NO2(g) + O2(g)
a. How many moles of N2O5 were used if 210g of NO2 were produced?
2N2O5(g) → 4NO2(g) + O2(g)
? moles
210g
Units match
210 g NO2
= 2.28
moles N2O5
b. How many grams of N2O5 are needed to produce 75.0 grams of O2?
2N2O5(g) → 4NO2(g) + O2(g)
75.0 g
? grams
75.0 g O2
= 506 grams N2O5
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Gram to Gram Conversions
Aluminum is an active metal that when placed in hydrochloric acid produces
hydrogen gas and aluminum chloride. How many grams of aluminum
chloride can be produced when 3.45 grams of aluminum are reacted with an
excess of hydrochloric acid?
2 Al(s) + 6HCl(aq)  2AlCl3(aq) +
3 H2(g)
First write a balanced
equation.
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Gram to Gram Conversions
Aluminum is an active metal that when placed in hydrochloric acid produces
hydrogen gas and aluminum chloride. How many grams of aluminum
chloride can be produced when 3.45 grams of aluminum are reacted with an
excess of hydrochloric acid?
2 Al(s) + 6HCl(aq)  2AlCl3(aq) +
? grams
3.45 g
3 H2(g)
Now let’s get organized.
Write the information
below the substances.
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gram to gram conversions
Aluminum is an active metal that when placed in hydrochloric acid produces
hydrogen gas and aluminum chloride. How many grams of aluminum
chloride can be produced when 3.45 grams of aluminum are reacted with an
excess of hydrochloric acid?
2 Al(s) + 6HCl(aq)  2AlCl3(aq) +
? grams
3.45 g
3 H2(g)
Units match
3.45 g Al
=
We
Now
Now
Let’s
must
use
use
work
the
always
thethe
molar
molar
problem.
convert
mass
ratio.to
convert
moles.
to grams.
17.0 g AlCl3
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Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
Nowthe
place
First copy down
numerical
the BALANCED the
information below
equation!
the compounds.
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Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
0.15 mol
? moles
0.10 mol
Hide
one
Two starting
amounts?
Where do we
start?
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Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
0.15 mol
? moles
Hide
0.10
mol
Based on:
0.15 mol KO2
KO2
= 0.1125 mol O2
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Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
0.15
mol
? moles
0.10 mol
Hide
Based on:
0.15 mol KO2
KO2
Based on: 0.10 mol H2O
H2O
= 0.1125 mol O2
= 0.150 mol O2
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Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
Determine the limiting reactant.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
0.15 mol
? moles
0.10 mol
Based on:
0.15 mol KO2
KO2
Based on:
H2O
= 0.1125 mol O2
It was limited by the
amount of KO2.
0.10 mol H2O
= 0.150 mol O2
H2O = excess (XS) reactant!
What is the theoretical yield?
Hint: Which is the smallest
amount? The is based upon the
limiting reactant?
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Theoretical yield vs. Actual yield
Suppose the theoretical yield for an
experiment was calculated to be
19.5 grams, and the experiment
was performed, but only 12.3
grams of product were recovered.
Determine the % yield.
Theoretical yield = 19.5 g based on limiting reactant
Actual yield = 12.3 g experimentally recovered
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Limiting/Excess Reactant Problem with % Yield
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O,
how many grams of O2 can be produced?
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
?g
120.0 g
47.0 one
g
Hide
Based on: 120.0 g KO2
KO2
= 40.51 g O2
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Limiting/Excess Reactant Problem with % Yield
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O,
how many grams of O2 can be produced?
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
?g
120.0
47.0 g
Hideg
Based on: 120.0 g KO2
KO2
= 40.51 g O2
Based on: 47.0 g H2O
H2O
= 125.3 g O2
Question if only 35.2 g of O2 were recovered, what was the percent yield?
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If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O,
how many grams of O2 can be produced?
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
?g
120.0 g
47.0 g
Based on: 120.0 g KO2
KO2
= 40.51 g O2
Based on: 47.0 g H2O
H2O
= 125.3 g O2
Determine how many grams of Water were left over.
The Difference between the above amounts is directly RELATED to the XS H2O.
125.3 - 40.51 = 84.79 g of O2 that could have been formed from the XS water.
84.79 g O2
= 31.83 g XS H2O
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