Unit VI – Stoichiometry
Applying Mole Town to Reactions
Learning Goals
• I can apply mole town to reactions to determine the amount
of product based on the amount of a reactant.
• I can apply mole town to reaction to determine the amount
of reactant needed to produce an amount product.
• I can calculate the molarity of a solution.
• I can determine how much of a known solution will be
needed to make a certain amount of another solution.
• I can determine the pH of an unknown acid or base based
upon the amount of a known acid or base that is necessary
to neutralize it.
• I can determine the limiting reagent in a reaction and how
much product will be produced based on that limiting
reagent.
Stoichiometry
• The study of quantities of materials consumed
and produced in chemical reactions.
– Simply the math behind chemistry.
• Given enough information, one can use
stoichiometry to calculate masses, moles, and
percents within a chemical equation.
Stoichiometry and Mole Town
• Stoichiometry adds a step to Mole Town
• When doing calculations to Mole Town, you have to
convert from one substituent in a reaction to
another in the same reaction
• Steps
1.
2.
3.
4.
Make sure the reaction you are using is balanced.
From wherever you are, get into moles (like before).
Switch between compounds using their molar ratios.
Get out of moles, if necessary to new unit of new
substituent.
Examples
23g Na
1 mole Na 2 mole NaCl 48.44g NaCl
22.99g Na 2 mole Na 1mole NaCl
• If 23 grams of Na are reacted with excess Cl2, how
many grams of NaCl will be produced?
1. Balance the equation
•
•
Na + Cl2  NaCl
2Na + Cl2  2NaCl
2. Grams  moles
3. Switch between compounds
1.
2 NaCl : 2 Na
4. Get back out of moles, into grams
= (23 x 1 x 2 x 48.44) / (22.99 x 2 x 1)
= 48.65 g of NaCl
Example 2
• If 44.8L of O2 are reacted with excess H2, how
many moles of H2O will be produced?
– H2 + O 2  H2 O
Molarity
• Most of chemistry happens in
solution
• Molarity is a measurement of
concentration of a solution
– The amount of a solute (the stuff
being dissolved) per liter of solution
• M = moles of a solute /liter of solution
Molarity without Stoichiometry
• Determine the moles of the solute (if
necessary)
• Divide by the volume
• Example:
– What is the molarity of 1.75 moles of solute
dissolved in 5.26L of solvent?
Molarity Examples
• What is the molarity of a solution when 2.78
moles of solute are dissolved in 265.2 mL of
solvent?
• How many grams of NaCl should be used to
prepare a 3.3L solution of saltwater with the
molarity of 4.26 M?
Molarity with Stoichiometry
• When using stoich to calculate molarity, always
start with the compound that the question is
NOT about.
1. Balance the equation
2. Start with the known compound
• Example: What mass of NaHCO3 is necessary to
neutralize 25.0L of 2.0M of HC2H3O2?
– NaHCO3 + HC2H3O2  NaC2H3O2 + H2O + CO2 = 4205g NaHCO3
25 L
HC2H3O2
2.0 mol
HC2H3O2
1.0 mol
NaHCO3
84.01 g
NaHCO3
= 25 x 2.0 x 1.0 x 84.01g
1 L HC2H3O2
1.0 mol
HC2H3O2
1.0 mol
NaHCO3
= 1 x 1.0 x 1.0
Example
• How many milliliters of 0.8M of acetic acid are
necessary to produce 450mL of carbon dioxide
gas at STP?
Example
• It took 10.0grams of sodium bicarbonate to
completely neutralize 15.0mL of acetic acid.
What was the molarity of the acid?
Warm Up
• How many milliliters of 0.6M hydrochloric acid
are needed to produce 60g of sodium chloride?
– HCl + NaOH  NaCl + H2O
Dilutions
• A solution with a known
concentration can be diluted with
water to prepare a solution with
any desired concentration that is
lower than the original solution.
– M1V1 = M2V2
•
•
•
•
M1 = the molarity of the original solution
V1 = the volume of the original solution
M2 = the molarity of the new solution
V2 = the volume of the new solution
Example
• Calculate the number of – Known?
• M1 = 12.0M
milliliters of
• M2 = 2.00M
concentrated
• V2 = 500mL
hydrochloric acid, 12.0M
– Unknown?
HCl, that must be used
• V1 = ?
to prepare 500. mL of – M V = M V
1 1
2 2
2.00M HCl solution.
• 12.0(V1) = 2.00(500mL)
• V1 = (1000 M x mL) / 12.0M
• V1 = 83.3 mL
Example
• Determine the number of milliliters of a 1.20M
concentrated hydrochloric acid solution needed
to prepare 2.00L of a 0.100M solution.
Example
• You wish to prepare 5.1M H2SO4 from stock
solution of sulfuric acid that is 18M. How many
milliliters of stock solution are necessary to
create 257mL of the 5.1M H2SO4?
Titrations
• REMEMBER: Acids release H+ when they dissolve
and bases release OH-.
• When acids and bases react with each other,
they neutralize.
– They create water and a salt.
• Acids have a pH below 7 and bases have a pH
above 7.
– pH measure the concentration of H+ ions in solution.
– Water has a pH of 7
Titrations
HALT… PLEASE have patience with yourself and
the process (and me ). I PROMISE it will
come if you TRY!!!!!!!!!!!!
Understanding something will be never come by whining
about how hard it is!!!
• Neutralizing an acid with a base to determine
the concentration of one or the other
• By knowing the concentration (molarity) of one
of the substances, you can calculate the
molarity of the other by doing a titration
• This process is bringing it all together
– Stoichiometry (ratios of reactions)
– Dilutions (molarity)
Titration Calculation Steps
1. Write down everything you know.
2. Write the reaction (or as much as you know)
– Make sure the reaction is balanced
3. Check all your info for consistency
– All volumes should be in L not mL, etc
4. Calculate the number of moles of the reactant
you know the most about
5. From the balanced equation, determine the mole
ratios
6. Calculate the number of moles of the unknown.
7. Using the volume of the unknown and moles of
unknown, calculate the Molarity
Titrations Helpful Hints
• Step One –
• Read the question carefully and ask yourself,
“What is the question giving me and what is it
asking for?”
• NEXT.. Figure out what tools you have to get
from one thing to the other?
• Next… read the question again!!
Example - 10.0mL of HCl (hydrochloric acid) are
titrated with 15.0mL of 2.0M sodium hydroxide
(NaOH). What is the molarity of the acid?
1. Known? 
•
•
•
reactant
5. Mole ratios
VHCl = 10.0mL
VNaOH = 15.0mL
MNaOH = 2.0M
•
2. Reaction?
•
HCl + NaOH  NaCl +
H2O
3. mL  L
•
•
1:1
6. Moles of Unknown?
7. Volume of unknown 
molarity of unknown
= 2.0 x 0.0150 x 1 mol
VHCl = 10.0mL = 0.0100 L
VNaOH = 15.0mL = 0.0150 L
= 0.03 / 0.01 = 3M
= 1 x 1 x 0.0100 L
4. Moles of known
2.0 moles NaOH
1L NaOH
0.0150 L NaOH
1 mol HCl
1 mol NaOH
0.0100 L HCl
Example
• If 25.6mL of 4.40M NaOH was required to
neutralize 325mL of H2SO4, what was the
molarity of the H2SO4 solution?
Example
• Calculate the molarity of a NaOH solution if a
31.22mL sample of it requires 12.53 mL of
0.100M H2SO4 to neutralize the base.
Warm Up
• If 25mL of 2.5M HCl are needed to neutralize
78mL of an unknown NaOH solution, what is the
molarity of the NaOH?
Limiting Reagent
• In a reaction, the reactant that limits the amount
of product formed is considered the limiting
reagent.
– That reactant must be completely used up and the other
reactant will be in excess
• Example: How many hamburgers can I make with 6
pieces of bread and 6 hamburger patties?
– When I consider just the bread –
• 6 slices │1 burger │ = 6 burgers = 3 burgers
•
│2 slices │ = 2
– When I consider just the patties –
• 6 patties│1 burger │ = 6 burgers = 1 burgers
•
│1 patty │ = 1
– The bread is the limiting reagent
Limiting Reagent Problems
1. You’ll be given 2 starting amounts.
2. Figure out how much of the desired product
can be formed in the right unit.
3. The limiting reagent limits the amount of
product that can be formed.
4. The answer is the minimum amount of product
formed!!
Example - Large quantities of ammonia, NH3, are burned
in the presence of a platinum catalyst to give nitrogen
monoxide. Suppose a vessel contains 2.20 moles of NH3
and 1.20 moles of O2. Which is the limiting reagent? How
many moles of NO could be obtained?
• 4 NH3 + 5O2  4 NO + 6 H2O
• Limiting reagent?
1. Known?
–
–
•
2.20 Moles of NH3
1.20 moles of O2
How much product?
2.20 mol NH3
4 mol NO
• O2
• How much NO?
• 0.96 mol
= 2.20 mol NO
4 mol NH3
1.20 mol NH3
4 mol NO
5 mol NH3
= 0.96 mol NO
Example
• Carbon disulfide, CS2, burns in oxygen. How
many liters of CO2 are there produced when
23.65 grams of CS2 and 32.34 grams of O2 react?
Warm Up
• If 25.2g of CO2 react with water, how much
H2CO3 will be produced? CO2 + H2O  H2CO3
• How many grams of NaCl are required to
produce 500mL of 2.2M solution?
• How much stock solution, 15M, is necessary to
produce 500mL of 2.5M solution?
• If 50mL of NaOH are required to neutralize
27mL of 0.5M HCl, what is the molarity of the
NaOH? HCl + NaOH  H2O + NaCl