STOICHIOMETRY TUTORIAL

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STOICHIOMETRY TUTORIAL

• Paul Gilletti

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Instructions This is a work along tutorial.
Each time you click the mouse or touch the space
bar on your computer, one step of the problem
solving occurs. Pressing the PAGE UP key will
backup the steps.
Get a pencil and paper, a periodic table and a
calculator, and lets get to work.
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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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Table of Contents Click on each tab to view
problem types.
View Complete Slide Show
Sample problem 1
Sample problem 2
Converting grams to moles
Mole to Mole Conversions
Gram-Mole and Gram-Gram Problems
Solution Stoichiometry Problems
Limiting/Excess/ Reactant and Theoretical Yield
Problems
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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
Units match
3. Convert using factors (ratios).
10 miles

inches
633600
Goal
Given
Convert
Menu
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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 k 1 x 103 c 1 x 10-2 1 hr 60
min 1 min 60 s
Goal
Given
km s
0.020
This is the same as putting k over k
c cancels c m remains
Units Match!
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Converting grams to moles.
Determine how many moles there are in 5.17 grams
of Fe(C5H5)2.
Goal
Given
units match
5.17 g Fe(C5H5)2
moles Fe(C5H5)2
0.0278
Use the molar mass to convert grams to moles.
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 Ba(OH)2 ? 2H2O BaCl2
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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
moles NO2
8.6
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
mole O2
2.2
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gram ? mole and 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)
210g
? moles
Units match
210 g NO2
moles N2O5
2.28
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
grams N2O5
506
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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?
Al(s) HCl(aq) ? AlCl3(aq)
H2(g)
2 6 2
3
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?
Al(s) HCl(aq) ? AlCl3(aq)
H2(g)
2 6 2
3
? grams
3.45 g
Now lets 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?
Al(s) HCl(aq) ? AlCl3(aq)
H2(g)
2 6 2
3
? grams
3.45 g
Units match
3.45 g Al
g AlCl3
17.0
Lets work the problem.
We must always convert to moles.
Now use the molar ratio.
Now use the molar mass to convert to grams.
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Molarity
Molarity is a term used to express concentration.
The units of molarity are moles per liter (It is
abbreviated as a capital M)
When working problems, it is a good idea to
change M into its units.
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Solutions
A solution is prepared by dissolving 3.73 grams
of AlCl3 in water to form 200.0 mL solution. A
10.0 mL portion of the solution is then used to
prepare 100.0 mL of solution. Determine the
molarity of the final solution.
What type of problem(s) is this?
Molarity followed by dilution.
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Solutions
A solution is prepared by dissolving 3.73 grams
of AlCl3 in water to form 200.0 mL solution. A
10.0 mL portion of the solution is then used to
prepare 100.0 mL of solution. Determine the
molarity of the final solution.
1st
3.73 g
mol L
0.140
200.0 x 10-3 L
molar mass of AlCl3
dilution formula
M1V1 M2V2
2nd
(0.140 M)(10.0 mL) (? M)(100.0 mL)
final concentration
0.0140 M M2
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Solution Stoichiometry
50.0 mL of 6.0 M H2SO4 (battery acid) were
spilled and solid NaHCO3 (baking soda) is to be
used to neutralize the acid. How many grams of
NaHCO3 must be used?
H2SO4(aq) 2NaHCO3 ? 2H2O(l) Na2SO4(aq)
2CO2(g)
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Solution Stoichiometry
50.0 mL of 6.0 M H2SO4 (battery acid) were
spilled and solid NaHCO3 (baking soda) is to be
used to neutralize the acid. How many grams of
NaHCO3 must be used?
H2SO4(aq) 2NaHCO3 ? 2H2O(l) Na2SO4(aq)
2CO2(g)
50.0 mL
? g
Our Goal
6.0 M

Look! A conversion factor!
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Solution Stoichiometry
50.0 mL of 6.0 M H2SO4 (battery acid) were
spilled and solid NaHCO3 (baking soda) is to be
used to neutralize the acid. How many grams of
NaHCO3 must be used?
H2SO4(aq) 2NaHCO3 ? 2H2O(l) Na2SO4(aq)
2CO2(g)
50.0 mL
? g
Our Goal
6.0 M

NaHCO3 2 mol
H2SO4 50.0 mL
NaHCO3 84.0 g
50.4
g NaHCO3
mol NaHCO3
1 mol H2SO4
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Solution Stoichiometry
Determine how many mL of 0.102 M NaOH solution
are needed to neutralize 35.0 mL of 0.125 M H2SO4
solution.
2 1
2 1
____NaOH ____H2SO4 ? ____H2O
____Na2SO4
First write a balanced Equation.
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Solution Stoichiometry
Determine how many mL of 0.102 M NaOH solution is
needed to neutralize 35.0 mL of 0.125 M H2SO4
solution.
2 1
2 1
____NaOH ____H2SO4 ? ____H2O
____Na2SO4
0.102 M
35.0 mL
? mL
Our Goal
Since 1 L 1000 mL, we can use this to save on
the number of conversions
Now, lets get organized. Place numerical
Information and accompanying UNITS below each
compound.
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Solution Stoichiometry
Determine how many mL of 0.102 M NaOH solution is
needed to neutralize 35.0 mL of 0.125 M H2SO4
solution.
2 1
2 1
____NaOH ____H2SO4 ? ____H2O
____Na2SO4
0.102 M
35.0 mL
? mL
Units Match
shortcut
H2SO4 35.0 mL
H2SO4 0.125 mol 1000 mL
H2SO4
NaOH 2 mol 1 mol H2SO4
1000 mL NaOH 0.102 mol NaOH
mL NaOH
85.8
Now lets get to work converting.
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Solution Stoichiometry
What volume of 0.40 M HCl solution is needed to
completely neutralize 47.1 mL of 0.75 M Ba(OH)2?
1st write out a balanced chemical equation
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Solution Stoichiometry
What volume of 0.40 M HCl solution is needed to
completely neutralize 47.1 mL of 0.75 M Ba(OH)2?
2HCl(aq) Ba(OH)2(aq) ? 2H2O(l)
BaCl2
47.1 mL 0.75 M
0.40 M
? mL
Units match
HCl 2 mol
Ba(OH)2 47.1 mL
HCl 1000 mL
176
mL HCl
0.40 mol HCl
1 mol Ba(OH)2
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Solution Stochiometry Problem
A chemist performed a titration to standardize a
barium hydroxide solution. If it took 23.28 mL
of 0.135 M hydrochloric acid to neutralize 25.00
mL of the barium hydroxide solution, what was the
concentration of the barium hydroxide solution in
moles per liter (M)?
2 1
2 1
____HCl(aq) ____Ba(OH)2(aq) ?
____H2O(l) ____BaCl2(aq)
25.00 mL
23.28 mL
? mol L
0.135 mol L
First write a balanced chemical reaction.
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Solution Stochiometry Problem
A chemist performed a titration to standardize a
barium hydroxide solution. If it took 23.28 mL
of 0.135 M hydrochloric acid to neutralize 25.00
mL of the barium hydroxide solution, what was the
concentration of the barium hydroxide solution in
moles per liter (M)?
2 1
2 1
____HCl(aq) ____Ba(OH)2(aq) ?
____H2O(l) ____BaCl2(aq)
25.00 mL
23.28 mL
Units match on top!
? mol L
0.135 mol L
mol Ba(OH)2 L
Ba(OH)2
0.0629
25.00 x 10-3 L Ba(OH)2
Units Already Match on Bottom!
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Solution Stochiometry Problem
48.0 mL of Ca(OH)2 solution was titrated with
19.2 mL of 0.385 M HNO3. Determine the molarity
of the Ca(OH)2 solution.
We must first write a balanced equation.
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Solution Stochiometry Problem
48.0 mL of Ca(OH)2 solution was titrated with
19.2 mL of 0.385 M HNO3. Determine the molarity
of the Ca(OH)2 solution.
Ca(OH)2(aq) HNO3(aq) ?
H2O(l)
2
Ca(NO3)2(aq)
2
48.0 mL
19.2 mL
? M
0.385 M
HNO3
19.2 mL
mol(Ca(OH)2) L (Ca(OH)2)

0.0770
48.0 x 10-3L
units match!
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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)
First copy down the the BALANCED equation!
Now place numerical the information below 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
0.10 mol
Hide
Based on KO2
0.15 mol KO2
mol O2
0.1125
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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 KO2
0.15 mol KO2
mol O2
0.1125
0.10 mol H2O
Based on H2O
mol O2
0.150
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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 KO2
0.15 mol KO2
mol O2
0.1125
It was limited by the amount of KO2.
0.10 mol H2O
Based on H2O
mol O2
0.150
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 g
Hide one
120.0 g KO2
Based on KO2
g O2
40.51
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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 g
Hide
Based on KO2
120.0 g KO2
g O2
40.51
47.0 g H2O
Based on H2O
g O2
125.3
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
120.0 g KO2
Based on KO2
g O2
40.51
47.0 g H2O
Based on H2O
g O2
125.3
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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Try this problem (then check your answer)
Calculate the molarity of a solution prepared by
dissolving 25.6 grams of Al(NO3)3 in 455 mL of
solution.
After you have worked the problem, click here to
see setup answer
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