Lesson: Calculations
Consider:
mol ratio
C (s)
+
1 atom
5 atoms
6.02 x 10 23 atoms
1 mol
1: 2: 1
2 S (s)

2 atoms
10 atoms
2(6.02 x 10 23 atoms)
2 mol
CS2 (g)
1 molecule
5 molecules
6.02 x 10 23 atoms
1 mol
In a balanced equation, the mole ratio corresponds to the coefficients.
1. Determine the mole ratio
a. N2 (g) + 3H2 (g)  2NH3 (g)
b.
If there are 4.0 moles of N2 how many moles of H2 are required?
2. Given: 4 NH3 (g) + 5 O2 (g)  4NO (g) + 6 H2O (g)
Mole ratio
How many moles oxygen are required to react with 3.46 x 10 -1 mol ammonia (NH3)?
3. Calculate the mass of oxygen needed to react with 5.00g of hydrogen gas given:
2H2 (g) + O2 (g)  2 H2O (l)
Mole ratio
a. # mol H2
b. # mol O2
c.
mass O2
MASS A  MOLES A  MOLES B  MASS B
2. What mass of hydrogen gas will be produced from 7.50 g of hydrochloric acid with excess
aluminum?
2 Al (s) + 6 HCl (aq)  2 AlCl3 (aq) + 3 H2 (g)
Mole ratio
a. # mol HCl
b. mol AlCl3
c. mass H2
Homework: pg. 298 (starting) #1-30
Lesson: Mass and Particle Stoichiometry
Recall: Avogadro’s Constant (NA)- relates # particles to the mole
6.02 × 1023 atoms / 1 mol } for element only
6.02 × 1023 molecules / 1 mol } for covalent compound
6.02 × 1023 formula units / 1 mol } for ionic compound
-
stoichiometry is the science of determining the relative ratios in which substances react
stoichiometric problems can be expanded to solve for the number of particles in a substance
Ex. According to the following equation: S8(s) + 12 O2(g)  8 SO3(g)
How many atoms of oxygen will form when 2.50 g of S 8(s) is reacted with excess oxygen gas?
S8(s) +
2.50 g
12 O2(g)

8 SO3(g)
#O atoms=?
# molecules SO3
# mol S8 --------------------------------------------> # mol SO3
In four separate steps:
In one step:
Homework: page 305 #10, worksheet
Stoichiometry Problems
1. Given the following equation: 2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O, show what the following molar ratios
should be.
a. C4H10 / O2
b. O2 / CO2
c. O2 / H2O
d. C4H10 / CO2
e. C4H10 / H2O
2. Given the following equation: 2 KClO3 ---> 2 KCl + 3 O2
How many moles of O2 can be produced by letting 12.00 moles of KClO3 react?
3. Given the following equation: 2 K + Cl2 ---> 2 KCl
How many grams of KCl is produced from 2.50 g of K and excess Cl2. From 1.00 g of Cl2 and excess K?
4. Given the following equation: Na2O + H2O ---> 2 NaOH
How many grams of NaOH is produced from 1.20 x 102 grams of Na2O? How many grams of Na2O are
required to produce 1.60 x 102 grams of NaOH?
5. Given the following equation: 8 Fe + S8 ---> 8 FeS
What mass of iron is needed to react with 16.0 grams of sulfur? How many grams of FeS are produced?
6. Given the following equation: 2 NaClO3 ---> 2 NaCl + 3 O2
12.00 moles of NaClO3 will produce how many grams of O2? How many grams of NaCl are produced when
80.0 grams of O2 are produced?
7. Given the following equation: Cu + 2 AgNO3 ---> Cu(NO3)2 + 2 Ag
How many moles of Cu are needed to react with 3.50 moles of AgNO3? If 89.5 grams of Ag were produced,
how many grams of Cu reacted?
8. Molten iron and carbon monoxide are produced in a blast furnace by the reaction of iron(III) oxide and coke
(pure carbon). If 25.0 kilograms of pure Fe2O3 is used, how many kilograms of iron can be produced? The
reaction is: Fe2O3 + 3 C ---> 2 Fe + 3 CO
9. The average human requires 120.0 grams of glucose (C6H12O6) per day. How many grams of CO2 (in the
photosynthesis reaction) are required for this amount of glucose? The photosynthetic reaction is: 6 CO 2 + 6
H2O ---> C6H12O6 + 6 O2
10. Given the reaction: 4 NH3 (g) + 5 O2 (g) ---> 4 NO (g) + 6 H2O (l)
When 1.20 mole of ammonia reacts, the total number of moles of products formed is:
a. 1.20 b. 1.50 c. 1.80 d. 3.00 e. 12.0
Lesson: Stoichiometry with Gases
Recall: Law of Combining Gas Volumes
- when gases react, the volumes of the reactants and products (measured at equal P and T) are
always in whole number ratios
- the mole ratios are the same as the ratios of the volumes of the gas
Ex. N2 (g) + 3H2 (g)  2NH3 (g)
If 15.0 L of H2 (g) reacts, calculate
a) the volume of N2 (g) used
b) the volume of NH3 (g) produced
a) vol N2 (g) used =
b) vol NH3 (g) produced =
Using the Ideal Gas Equation (PV=nRT) in Stoichiometry
-
once the # mol of a gas is found it can be converted into the units required by using the set
conditions in the ideal gas equation
Ex. Fe(s) + H2SO4(aq)  H2(g) + FeSO4(aq)
What volume of H2(g) is produced when 0.164 kg of Fe(s) reacts with excess sulphuric acid at SATP?
# mol Fe =
#mol H2 =
n=
P=
V= ?
R
T=
PV = nRT
V=
.:
L of H2 is produced by this reaction
This same process can be used to find the mass of a solid reactant or product if the volume of a gas
involved in the reaction is known.
Lesson: Stoichiometry with Solutions
Many different types of problems involving solution stoichiometry are possible. All solution problems
involve some form of the equation:
# mol = conc in mol/L × vol in L
Ex. Calculate the volume of 2.00 mol/L silver nitrate solution that is needed for 12.0 g of copper metal
to react according to the following equation:
Cu(s) + 2 AgNO3(aq)  Cu(NO3)2(aq) + 2 Ag(s)
Ex. When a solution of silver nitrate is added to a solution of calcium chloride the following reaction
takes place:
2AgNO3(aq) + CaCl2(aq)  2 AgCl(s) + Ca(NO3)2(aq)
Calculate the volume of 0.105 mol/L silver nitrate that must be added to 25.0 ml of 0.255 mol/L
calcium chloride solution in order to produce the maximum mass of silver chloride.
2AgNO3(aq) + CaCl2(aq)  2 AgCl(s) + Ca(NO3)2(aq)
More problems to try:
1. Aluminum reacts with hydrochloric acid according to the equation:
2Al(s) + 6 HCl(aq)  2 AlCl3(aq) + 3H2(g)
Calculate the volume of 1.50 mol/L hydrochloric acid that is required for 5.40 g of aluminum to react
completely.
2. Zinc metal reacts with hydrochloric acid according to the equation:
Zn(s) + 2 HCl(aq)  H2(g) + ZnCl2(aq)
A piece of zinc metal is found to require 75.0 mL of 3.00 M HCl for a complete reaction. Determine
the mass of the piece of zinc metal.
3. Calculate the volume of 0.110 mol/L sodium sulphate required to precipitate the max mass of
barium sulphate from 60.0 mL of 0.145 mol/L barium chloride solution.
Homework: page 560 #31-40, page 417 #11-19
Lesson: Limiting Reagent
Recall: Stoichiometry with Solutions
-mixing two solutions in stoichiometric amounts to produce the maximum amount of precipitate
-what if the reactants were not present in the exact mole ratios?
The Limiting Reactant (Reagent)
-see lettuce example pg. 296
-usually in chemical reactions, one reactant is in excess, and the other reactant limits how much
product can be made
-the reactant that is completely used up in the reaction is called the limiting reactant
-the reactant that is left over at the end of the reaction is called the excess reactant
-when the limiting reactant is completely used up the reaction stops and no more product can be
formed no matter how much excess reactant is left over
Identifying the Limiting Reactant in Stoichiometric Problems
-unless you are told which reactant is in excess you must find the limiting reactant before performing
a stoichiometric calculation
Ex. If 4.87 g of litihium nitride reacts with 5.80 g of water
a) find the limiting reactant
b) what volume of ammonia gas is formed if performed at 21ºC, 102 kPa?
Step 1: Write the balanced equation
Step 2: find the # mols of each reactant
#mol Li3N
=
# mol H20
=
Step 3: calculate the # mol of product that can be produced by each reactant and determine which
one produces the smaller amount of product (this is the limiting reactant)
# mol NH3 (if Li3N limiting) =
# mol NH3 (if H20 limiting) =
is the limiting reactant
Step 4: calculate the amount of product as specified in the question
P= 102 kPa
V=?
n = 0.107 mol
R = 8.31 L·kPa/mol·K
T= 21ºC =
K
PV = nRT
Homework: starts pg. 309 #31-49
Lesson: Percentage Yield
The amount of product expected from a reaction rarely equals the amount that is actually produced.
The Percentage Yield indicates the efficiency of a reaction. In the chemical industry and
pharmaceuticals this is very important as reactants are very expensive.
Theoretical yield
- calculated from a balanced equation
- “what you should get”
Actual yield
- the amount of product obtained from an experiment (reaction)
- “what you actually got”
Actual yield is usually less than theoretical yield because:
1. experimental error (spillage); poor technique
2. other products are formed during the reaction (by-products)
3. reactants are impure
% yield of product =
actual yield ____
theoretical yield
x 100 %
E.g.:
1. Ammonia is prepared by reacting nitrogen gas with hydrogen gas. When 750 g of nitrogen reacts
with enough hydrogen, the theoretical yield of ammonia is 9.10g. If 1.72 g of ammonia is actually
obtained, calculate the percent yield given:
N2 (g) + 3 H2 (g)  2 NH3 (g)
Actual yield NH3 = 1.72 g
Theoretical yield = 9.10g
% yield NH3 =
Note: You can calculate (predict) the actual yield if you know the % yield.
2. The theoretical yield of MgCO3 is 25.8 g. If the % yield of the MgCO3 in a reaction is 83.5 %, what
amount of MgCO3 will be obtained?
actual mass MgCO3 =
Now we are able to:
a) predict theoretical yield using stoichiometric calculations and the limiting reactant
b) use experimental results and theoretical yield to find % yield
1) Solid lead (II) nitrate decomposes upon heating to give solid lead (II) oxide, nitrogen dioxide gas
and oxygen gas. When 3.31 g of lead (II) nitrate was heated in an experiment, the lead (II) oxide was
found to weigh 1.75g. Calculate the % yield obtained. Answer 78.5 %
2) When heated, sodium metal combines with chlorine gas to form solid sodium chloride. If 8.30g of
sodium and 14.0g of chlorine are heated together a total of 19.5g of sodium chloride is isolated.
Determine the % yield of the sodium chloride.
(Hint: Steps are 1) balanced equation 2) find LR 3) find theoretical yield 4) find % yield
Homework: page 319 #51-60
Lesson: Neutralization Reactions
Acid-Base Reactions
Recall:
Acid - dissociates in water to form H+ (Arrhenius)
- proton donor (Bronsted-Lowry)
Base - dissociates in water to form OH- (Arrhenius_
- proton acceptor (Bronsted-Lowry)
The reaction between an acid and a base is a neutralization reaction.
Acid + Base  Salt + Water
-
a salt is an ionic compound containing an anion from an acid and cation from a base
ex. HCl(aq) + KOH(aq)  KCl(aq) + H2O(l)
What will the pH be?
-
when any strong base reacts with any strong acid in the mole ratio from the balanced equation, a
neutral solution of a salt is formed
How could we know a neutralization reaction occurred?
Acid-Base Indicators
- weak, monoprotic acid that changes colour in acidic and basic conditions
ex. Phenolphthalein is colourless in acidic, pink in basic
Stoichiometric Calculations
- A Titration is an experimental procedure where the concentration of one solution is determined by
quantitatively observing its reaction with a solution of known concentration (standard solution).
The goal of a titration is to find a point at which the # mol of standard solution is stoichiometrically
equal to the original # mols of unknown solution. This is the equivalence point where all of the mol
H+ in one solution have reacted with an equal # mol OH- from the other solution.
A buret is used to add solution
drop by drop
A transfer pipet is used to measure a
precise volume of standard solution
into a flask (containing a few drops of indicator)
The end-point of a titration is when the indicator changes colour. An indicator is chosen so that its
end point is very close to the equivalence point.
Ex. In a titration, 31.24 mL (measured in a buret) of NaOH(aq) with a conc of 0.1004 mol/L was
required to neutralize 25.0 mL of HCl(aq). Calculate [HCl].
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
# mol NaOH =
# mol HCl =
[HCl] =
Demonstration- Titration
1. Rinsing the pipette- with water, then with solution
2. Rinse Erlenmeyer flask- with dH2O
3. Fill Pipette, add to flask
-
Add to 50 ml mark with dH2O
Add 2-3 drops of phenolphthalein indicator
4. Rinse buret- with water, then with solution (use funnel)
-check for air bubbles once filled
-allow to drain to 0.00ml or below
-remove funnel before taking initial reading
5. Titrate
-white paper under flask
-swirl continuously with one had while controlling buret tap with the other
-look for a permanent pale pink
-rinse drops in with dH2O when you are getting close to end point
-take your final reading
Homework: pg. 466 #1-10, read pg. 468-469, pg. 470 all q’s
The Limiting Reactant and Percentage Yield of a Chemical Reaction
Purpose:
To determine the limiting reactant in the chemical reaction between aluminum and copper (II) chloride dihydrate
and to determine the percent yield in this reaction.
Materials:
aluminum foil, approx. 6 cm x 10 cm
copper (II) chloride dihydrate
beakers, 250 mL & 600 mL
100 mL graduated cylinder
stirring rod
ring clamp
retort stand
funnel
distilled water
hot plate
beaker tongs
evaporating dish
filter paper
clay triangle
wire gauze
tweezers or crucible tongs
Procedure:
1. Fold the 6 cm x 10 cm piece of aluminum foil lengthways twice to make a 1 cm x 10 cm strip. Bend the strip into a circle.
Record the mass of the aluminum.
2.
Using the 250 mL beaker and the TARE button, measure 2 g of copper (II) chloride dihydrate. Record the accurate mass in the
data chart.
3.
Using a graduated cylinder, add 50 mL of distilled water. Stir until the copper (II) chloride dihydrate has dissolved.
4.
Add the aluminum strip to the solution.
5.
Set up the hot plate. Gently heat until the blue colour disappears or until all of the aluminum disappears. Do not touch the
aluminum during the reaction. If there is excess foaming, remove the beaker using the beaker tongs, until the foaming stops.
Continue to heat if necessary.
6.
Turn off the hot plate. Using the beaker tongs remove the beaker to the wire gauze. Allow the beaker to cool for 10 minutes.
While the beaker is cooling, continue with step 7 and 8.
7.
Determine the mass of the folded filter paper. Record the mass in the data chart.
8.
Set up a filtration using the filter paper, funnel, clay triangle and a 600 mL beaker.
9.
Once the beaker has cooled, use tweezers to dislodge the copper and remove the excess aluminum foil by shaking carefully (if
there was any).
10. Filter, making sure to transfer all of the copper pieces to the filter.
11. Use the distilled water bottle to rinse the copper through the filter paper.
12. When the solution has completely run through the funnel, carefully remove the filter paper from the funnel and place it on the hot
plate.
13. Allow the filter paper to dry. Make sure you watch it carefully so the paper does not burn!
14. Record the mass of the filter paper with the copper.
15. Dispose of all equipment as instructed by the teacher.
Background Calculations:
1.
2.
Write the balanced chemical reaction including states.
Practice calculating the limiting reactant, theoretical yield of copper and percent yield using the following data:
Mass of Al = 3.30 g
Mass of CuCl2  2 H2O =2.23 g
Actual Yield of Cu = 0.56 g
Observations
Table 1 : Quantitative Observations of Reactants and Products
Table 2: Qualitative Observations of Reactants and Products
Name:_________________Stoichiometry Assignment 2014
T/I
/
Com
/8
Answer the following questions. Use proper form, units, significant figures. Show all of your work. Draw a box
around your final answer.
1. How many moles of oxygen are produced when 3.50 mol of water decomposes?
2. When iron rusts, iron metal reacts with oxygen gas to produce iron (III) oxide. How many grams of iron (III) oxide
would form if a 745 g iron hinge turned completely to rust?
3. Given the following equation: 4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
a) determine the number of molecules of oxygen gas that are required to completely react with 23.2 g of
ammonia
b) calculate the mass of ammonia that must react with oxygen gas to produce 8.00 × 10 23 molecules of nitrogen
monoxide
4. Phosphorus reacts with oxygen gas to form solid phosphorous (V) oxide. How many atoms of oxygen are
required to react with solid phosphorous to form 26.8 g of phosphorous (V) oxide?
5. Zinc reacts with hydrochloric acid. Calculate the percent yield of zinc chloride if 1.541 g of it is produced when
0.999g of hydrochloric acid reacts with excess zinc.
6. When heated, sodium metal reacts with chlorine gas to produce solid sodium chloride. If 8.30 g of sodium and
14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage
yield.
7. Solid lead(II)nitrate decomposes on heating to give solid lead(II)oxide, nitrogen dioxide gas, and oxygen gas.
When 3.31 g of lead(II) nitrate was heated in an experiment, there was 1.75 g of lead(II) oxide obtained.
Calculate the percentage yield for the experiment.