Transcript Stoichiometry - Mr Field's Chemistry Class

Stoichiometry
Mr Field
Using this slide show

The slide show is here to provide structure to the lessons, but not
to limit them….go off-piste when you need to!

Slide shows should be shared with students (preferable electronic to
save paper) and they should add their own notes as they go along.

A good tip for students to improve understanding of the
calculations is to get them to highlight numbers in the question and
through the maths in different colours so they can see where
numbers are coming from and going to.

The slide show is designed for my teaching style, and contains only
the bare minimum of explanation, which I will elaborate on as I
present it. Please adapt it to your teaching style, and add any notes
that you feel necessary.
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Menu:
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



Lesson 1 – The Mole
Lesson 2 – Moles and
Molar Mass
Lesson 3 - Solutions
Lesson 4 - Formulas
Lesson 5 - Equations





Lesson 6 – Theoretical
Yields
Lesson 7 – Molar Volume
of Gases
Lesson 8 – Ideal Gases
Lesson 9 - TEST
Lesson 10 – Test Debrief
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Lesson 1
The Mole and the Avogadro Constant
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Intro to Unit

Brainstorm everything you know about quantitative
chemistry (equations, moles and all that stuff!)
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Overview

Copy this onto an A4 page. You should add to it as a
regular review throughout the unit.
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Assessment

Test at the end of the unit (100%)

Approx lesson 9
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We Are Here
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Lesson 1: Meet the Mole

Objectives:

Understand the phrase ‘a mole’ as just a large number

Define ‘a mole’ according to Avogadro’s number

Complete an experiment to determine Avogadro’s number

Calculate numbers of particles using Avogadro’s number
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What do the following have in common?
A dozen
A grand
A score
A pair
A trio
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A Mole

A mole is a large number:

A mole is:

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

6.02 x 1023
602,000,000,000,000,000,000,000
Six hundred and two thousand quadrillion
Given the symbol, L
This number is called Avogadro’s number after the Italian scientist
Amedeo Avogadro who first proposed it
Picture – all the grains of sand, on all the beaches and in all the
deserts in the world; that is about a tenth of a mole!
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Some Calculations
•We can use this equation
to calculate a number of
moles from a number of
particles
N
n
L
•Where:
•n = quantity in moles
•N = number of
particles
•L = Avogadro’s
Constant (6.02x1023)
1.
Example 1: You have 3.01x1022 atoms of carbon. How
many moles is this?
•n = N / L
•n(C) = 3.01x1022 / 6.02x1023
•n(C) = 0.0500 mol
2.
Example 2: You have 6.02x1024 molecules of water.
How many moles of hydrogen atoms are present?
•n = N / L
•n(H) = 2 x n(H2O)
•n(H) = 2 x 6.02x1024 / 6.02x1023
•n(H) = 20.0
3.
Example 3: How many atoms of hydrogen are there in
2.5 moles of methane (CH4)?
•N(H) = 4 x N(CH4)
•N(H) = 4 x n(CH4) x L
•N(H) = 4 x 2.5 x 6.02x1023
•N(H) = 6.02x1024
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Some Calculations
1.
How many moles of ethene
(C2H4) are there in 1.5x1022
molecules?
2.
How many moles of oxygen
atoms in 1.20x1024 molecules of
sulphuric acid (H2SO4)? How
many moles of oxygen molecules
could they make?
3.
How many molecules in 3.0
moles of nitrogen gas?
4.
How many chloride ions are
released on dissolving 0.050
moles of calcium chloride
(CaCl2)?
BONUS QUESTION: How many moles of people are there on the planet?
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How do you determine Avogadro’s number?

There are a number of ways including:

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
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Electrolysis
Measurement of electron mass
X-ray crystal density
Molecular monolayers
You will be determining L by measuring the area of a
molecular monolayer

Not the most accurate but feasible in our lab!
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Key Points
N
n
L
particles
moles 
23
6.02 x10
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Lesson 2
Moles and Molar Mass
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Refresh

How many molecules are
present in a drop of ethanol,
C2H5OH, of mass 2.3 × 10–3
g? (L = 6.0 × 1023 mol–1)
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

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
A.
B.
C.
D.
3.0 × 10 19
3.0 × 1020
6.0 × 1020
6.0 × 1026


How many oxygen atoms are
there in 0.20 mol of ethanoic
acid, CH3COOH?
A. 1.2 × 1023
B. 2.4 × 1023
C. 3.0 × 1024
D. 6.0 × 1024
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Lesson 2: Moles and Molar Mass

Objectives:

Calculate the relative mass and molar mass of substances

Relate the mass of a substance to a quantity in moles

Conduct an experiment to determine the number of moles of
water of crystallisation
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Atomic and Formula Mass

Ar



Mr


is the relative atomic mass of an element
found in the periodic table
is the relative molecular mass of a compound
To calculate Mr, you add the Ar of all the atoms in a compound
1.
What does the term ‘relative’ mean?
2.
What determines the mass of an atom and how can Ar be a
decimal number?
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Calculating Mr

HCl

Ar(H) = 1.01
Ar (Cl) = 35.45

Mr = 1.01 + 35.45 = 36.46
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
C 2 H4


Ar(C) = 12.01
Ar (H) = 1.01

H2SO4

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Mr = 2x12.01 + 4x1.01
= 16.06
Mr = 2x1.01 + 32.06 + 4x16.00
= 98.08
Mg(OH)2


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
Ar(H) = 1.01
Ar (S) = 32.06
Ar(O) = 16.00

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Ar(Mg) = 24.31
Ar(O) = 16.00
Ar (H) = 1.01
Mr = 24.31 + 2x16.00 + 2x1.01
= 58.33
Calculate Mr for:

Br2

(NH4)2SO4

C3H8

C6H12O6
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Molar Mass, Mm

This is the mass of one mole of something.

To calulate Mm, simply stick ‘g’ for grams on the end of Mr.

For example:



Mr(H2O) = 18.02
Mm(H2O) = 18.02 g
Note: This is why the value of L was chosen to be what
it was.
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Relating ‘mass’ and ‘molar mass’

You need to be able to solve problems like:




How many moles of Y is mass X?
What is the mass of X moles of Y?
X moles of Y has a mass of Z, what is it’s molar mass?
Use this equation:
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For example:

How many moles of water are
present in 27.03g?



Calc Mm(H2O):
 Mm(H2O)= 2x1.01 + 16.00
= 18.02
Find n(H2O):
 n(H2O) = M / Mm
= 27.03 / 18.02
= 1.50 mol

What is the mass of 4.40 mol of
iron (III) oxide (Fe2O3)?


Calc Mm(Fe2O3):
 Mm(H2O)= 2x55.85 + 3x16.00
= 159.70 g
Find M(Fe2O3):
 M(Fe2O3) = n x Mm
= 4.40 / 159.70
= 703 g
1.30 moles of an unknown compound has a mass of 20.9g, what is it’s molar
mass?

Mm(unknown)= M / n
= 20.9 / 1.30
= 16.1 g/mol
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Questions
1.
Calculate the
mass of 0.10
mol of benzene
(C6H6)
2.
Calculate the
mass of 0.75
mol of
ammonium
nitrate
(NH4NO3)
3.
4.
What quantity
of iron (III)
oxide is present
in a 1.0 kg
sample?
What quantity
of cobalt (II)
chloride (CoCl2)
is present in a
2.40 g sample?
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5.
8.8 moles of a
compound has
a mass of 1.41
kg. Calculate
its molar mass.
6.
0.010 mol of
an oxide of
hydrogen has a
mass 0.340 g.
Deduce it’s
formula.
Moles of Water of Crystallisation

Many compounds can incorporate water into their crystal
structure, this is called the water of crystallisation.

CoCl2 is blue

The ‘.’ means the water is ‘associated’ with the CoCl2


CoCl2.6H2O is pink
It is loosely bonded, but exactly how is unimportant
In this experiment you will calculate the moles of water
of crystallisation of a compound
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Key Points
m ass
m oles
m olarm ass
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Lesson 3
Solutions
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Refresh

A student reacted some salicylic acid with excess ethanoic
anhydride. Impure solid aspirin was obtained by filtering the
reaction mixture. Pure aspirin was obtained by recrystallisation.
The following data was recorded by the student.


1.
Mass of salicylic acid used
Mass of pure aspirin obtained
3.15 ± 0.02 g
2.50 ± 0.02 g
Determine the amount, in mol, of salicylic acid,
C6H4(OH)COOH, used.
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Lesson 3: Solutions

Objectives:

Understand the relationship between concentration, volume
and moles

Prepare a standard solution of silver nitrate.

Pose and solve problems involving solutions (of the chemical
kind not the answers kind)
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Solutions Basics

Aqueous copper sulfate solution:
+
SOLUTE
SOLVENT
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SOLUTION
Concentration

This is the strength of a solution.
Most Concentrated
Least Concentrated
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Molarity

The number of moles of a substance dissolved in one litre of a
solution.
moles
concentrat ion 
volume

Units: mol dm-3




Pronounced: moles per decimetre cubed
Units often abbreviated to ‘M’ (do not do this in an exam!)
Volume must be in litres (dm3) not ml or cm3
This is the most useful measure of concentration but there are
others such as % by weight, % by volume and molality.
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Example 1:

25.0 cm3 of a solution of hydrochloric acid contains
0.100 mol HCl. What is it’s concentration?

Answer:

Concentration
= moles / volume
= 0.100 / 0.0250
= 4.00 mol dm-3

Note: the volume was first divided by 1000 to convert to dm3
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Example 2:

Water is added to 4.00 g NaOH to produce a 2.00 mol dm-3
solution. What volume should the solution be in cm3?

Calculate quantity of NaOH:


n(NaoH)
= mass / molar mass
= 4.00/40.0
= 0.100
Calculate volume of solution:

Volume
= moles / concentration
= 0.100 / 2.00
= 0.0500 dm3
= 50.0 cm3
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Example 3:

It is found by titration that 25.0 cm3 of an unknown solution of
sulfuric acid is just neutralised by adding 11.3 cm3 of1.00 mol
dm-3 sodium hydroxide. What is the concentration of sulfuric
acid in the sample.
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O

Use:
n 1 C1 V 1 = n 2 C2 V 2
1 x C1 x 25.0 = 2 x 1.00 x 11.3
Where:
n = coefficient
C = concentration
V = volume
‘1’ refers to H2SO4
‘2’ refers to NaOH
C1 = (2 x 1.00 x 11.3) / (1 x 25.0) = 0.904 mol dm-3
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Questions
1.
You have 75.0 cm3 of a 0.150 mol dm-3 solution of zinc sulphate (ZnSO4).
What mass of zinc sulphate crystals will be left behind on evaporation of
the water?
2.
What mass of copper (II) chloride (CuCl2) should be added to 240 cm3
water to form a 0.100 mol dm-3 solution?
3.
A 10.0 cm3 sample is removed from a vessel containing 1.50 dm3 of a
reaction mixture. By titration, the sample is found to contain 0.0053 mol
H+. What is the concentration of H+ in the main reaction vessel?
4.
The titration of 50.0 cm3 of an unknown solution of barium hydroxide
was fully neutralised by the addition of 12 cm3 of 0.200 mol dm-3
hydrochloric acid solution. What concentration is the barium hydroxide
solution?
Ba(OH)2 + 2 HCl  BaCl2 + 2 H2O
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Preparing a Standard Solution

Prepare standard solutions
silver nitrate with
concentrations of 0.25-0.75 mol
dm-3

You will use these in a future
lesson so make them well!
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Problem Time

Write three stoichiometry problems using any of the
ideas from the unit so far (make sure you know the
answer).

In ten minutes time you will need to give your problems
to a classmate to solve.
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Key Points
moles
concentrat ion 
volume
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Remember!
Alcohol is not the
answer but it is a
solution*!
*Of ethanol, water and various other bits and pieces
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Lesson 4
Formulas
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Refresh

Which statement about solutions is correct?
A.
B.
C.
D.

When vitamin D dissolves in fat, vitamin D is the solvent and fat is
the solute.
In a solution of NaCl in water, NaCl is the solute and water is the
solvent.
An aqueous solution consists of water dissolved in a solute.
The concentration of a solution is the amount of solvent dissolved
in 1 dm3 of solution
A student added 27.20 cm3 of 0.200 mol dm–3 HCl to 0.188 g
of eggshell. Calculate the amount, in mol, of HCl added.
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Lesson 4: Formulas

Objectives:

Understand the difference between empirical and molecular
formulas

Experimentally determine the empirical formula of a
compound.

Solve problems involving empirical and molecular formulas
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Formulas

You have one minute to write down as many different
chemical formulas as you can.

Go!
No!!!!

What is the job of a formula?
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Types of Formula

Empirical:


Molecular:


The ratio of the atoms of each element in a compound in its
lowest terms
The number of atoms of each element in a molecule
You will meet other types in the organic chemistry unit
including structural, displayed and skeletal.
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Examples

Oxygen





Molecular: O2
Empirical: O
Water


Molecular: C2H6
Empirical: CH3
Glucose



Molecular: H2O
Empirical: H2O
Ethane


Sodium Chloride



Molecular: C6H12O6
Empirical: CH2O
Molecular: n/a*
Empirical: NaCl
Copper Sulphate


Molecular: n/a*
Empirical: CuSO4
*Why do these two not have an empirical formula?
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Your turn…
Write empirical and molecular formulas for each of the following:

Chlorine gas

Sulphuric acid

Propane

Ethanol
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Determining empirical formulas from % by
mass data*

For example: a sample of a compound contains 20%
hydrogen and 80% carbon by mass.
C:
80%
6.67
1

H
20%
20
3
write out the elements as a ratio
write the % composition below
divide each % by the Ar of each element
divide each number by the smallest of the two numbers
Empirical formula = CH3
*This may seem like an odd thing to do but this data is
easily determined in the lab and so is very useful
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Determining Molecular Formulas From
Empirical Formulas

You need to know the molecular mass (can be found from mass
spectrometry (see Atomic Structure unit).

Following on from the previous question. Your unknown molecule is
found to have Mr = 30.06

Determine empirical formula mass:


Divide molecular mass by empirical formula mass to tell you the
multiplier for the empirical formula:


Mr(CH3) = 12.00 + (3 x 1.01) = 15.03
Multiplier = 30.06 / 15.03 = 2
Multiply the empirical formula by the multiplier:

Molecular formula = CH3 x 2 = C2H6
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Determining an empirical formula

You will be determining the empirical formula of an
unknown compound of potassium, chlorine and oxygen.

Follow the instructions here
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Practice Questions

Begin the practice questions found here

Complete for homework.
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Key points:

Empirical formula gives the lowest ratio of atoms

The molecular formula is a whole multiple of the
empirical formula

When using % by mass data, start by dividing each % by
the Ar of that element.
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Lesson 5
Equations
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Refresh

A toxic gas, A, consists of 53.8 % nitrogen and 46.2 % carbon
by mass and has a molar mass of 50.81 g/mol.

Determine the empirical formula of A.

Determine the molecular formula of A
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Lesson 5: Equations

Objectives:

Know how to construct and balance symbol equations

Apply the concept of the mole ratio to determine the amounts
of species involved in chemical reactions

Meet the idea of the ‘limiting reagent’
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Word Equations

Hydrogen
+
Oxygen

REACTANTS

Water
PRODUCTS
The arrow means: ‘makes’ or ‘becomes’ not ‘equals’
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Symbol Equations

2 H2 + O2  2 H2O
4H
4H
2O
2O

The bold numbers are called coefficients and tell you the
number of each molecule involved in the reaction




vs
H2 + O2  H2 O
2H
2H
2O
1O
Required to balance the equation
Without them the equation does not balance – each side of the
reaction would have different numbers of each atom – which would
break physics
You can’t change the little numbers in the formulas as this changes
the chemical
If there is no coefficient, it is ‘1’
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Tips for balancing equations

Only change the coefficients

Only make one change at a time

Re-count the numbers of each atom after each change

Try keeping a tally-chart of the numbers of atoms on each side of
the equation

If you get to a point where you just need a fractional amount of a
compound, write it in and then multiply everything by the
denominator of the fraction.

BE PATIENT!!!!
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Construct equations and then balance each
of the following:

Magnesium (Mg) reacts with hydrochloric acid (HCl) to make magnesium
chloride (MgCl2) and hydrogen gas

Ethane (C2H6) reacts with oxygen gas to make carbon dioxide and water

Lead nitrate (Pb(NO3)2) reacts with aluminium chloride (AlCl3) to make
aluminium nitrate (Al(NO3)3) and lead chloride (PbCl2)

Barium nitride (Ba3N2) reacts with water to make barium hydroxide
(Ba(OH)2) and ammonia (NH3)

Extension: Write a flow chart that enables you to balance equations
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Mole Ratios

This is the ratio of one compound to another in a balanced
equation.

For example, in the previous equation
2 H2 + O2  2 H2O

Hydrogen, oxygen and water are present in 2:1:2 ratio.

This ratio is fixed and means for example:



0.2 mol of H2 reacts with 0.1 mol of O2 to make 0.2 mol H2O
5 mol of H2 reacts with 2.5 mol of O2 to make 5 mol of H2O
To make 4 mol of H2O you need 4 mol of H2 and 2 mol of O2
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Mole Ratios cont….

2 Al(OH)3 + 3 H2SO4  Al2(SO4)3 + 3 H2O

In the above equation, what quantity of Al(OH)3 in moles is
required to produce 5.00 mol of H2O?

Divide the given moles by the given coefficient and multiply by
the target coefficient:

n(Al(OH)3) = 5.00/3 x 2 = 3.33 mol*
*Dividing by ‘3’ and multiplying by ‘2’ uses the H2O:Al(OH)3 mole
ratio of 3:2 to adjust the number of moles.
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The Limiting Reagent

In a reaction, there is we can describe reactants as being ‘limiting’ or in ‘excess’


Limiting – this is the reactant that runs out
Excess – the reaction will not run out of this
2 H2 + O2  2 H2O

For example, if you have 2.0 mol H2 and 2.0 mol O2



To determine this, divide the quantity of each reactant by its coefficient in
the equation. The smallest number is the limiting reactant:



H2 is the limiting reactant – it will run out
O2 is present in excess – there is more than enough
H2: 2.0 / 2 = 1.0 – smallest therefore limiting
O2: 2.0 / 1 = 2.0
You should use the limiting reactant when doing all further calculations including:


Determining amounts of products formed
Determining amounts of other reactants used
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Use the idea of limiting reactants to identify
the limiting reactant and solve the problem.

Mg + 2 HCl  MgCl2 + H2


2 C2H6 + 10 O2  4 CO2 + 6 H2O


0.55 mol C2H6 reacting with 3.50 mol O2. How many moles of CO2
is formed?
3 Pb(NO3)2 + 2 AlCl3  2 Al(NO3)3 + 3 PbCl2


0.15 mol Mg reacting with 0.25 mol HCl. How many moles of H2 is
formed?
2.60 mol Pb(NO3)2 reacting with 3.00 mol AlCl3. Which element is in
excess and how much remains after the reaction?
Ba3N2 + 6 H2O  3 Ba(OH)2 + 2 NH3

1.45 mol Ba3N2 reacting with 10.0 mol H2O. Which reactant is
present in excess and how much remains after the reaction?
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Key Points

Balance equations by playing with coefficients

Use mole ratios to work quantities of chemicals involved
in reactions

Divide the quantity of each reactant by its coefficient to
determine the limiting reactant
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Lesson 6
Theoretical Yields
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Refresh
Equal
masses of the metals Na, Mg, Ca and Ag are added to separate samples of excess
HCl(aq). Which metal produces the greatest total volume of H2(g)?
A.
Na
B.
Mg
C.
Ca
D.
Ag
Chloroethene, C2H3Cl, reacts
with oxygen according to the equation below.
2C2H3Cl(g) + 5O2(g) → 4CO2(g) + 2H2O(g) + 2HCl(g)
What is the amount, in mol, of H2O produced when 10.0 mol of C2H3Cl and 10.0 mol of
O2 are mixed together, and the above reaction goes to completion?
A.
4.00
B.
8.00
C.
10.0
D.
20.0
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Lesson 6: Theoretical Yields

Objectives:

Use the idea of limiting reactant to determine the proportions
of an unknown mixture

Use stoichiometry to calculate theoretical yields
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Preparation of Silver Chromate

In this experiment you will
prepare silver chromate

The challenge is that you
will start with an unknown
mixture of the reactants
and will have to use your
knowledge of
stoichiometry to work out
its composition
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Determining Theoretical Yield


The theoretical yield is the amount of product you expect to make if all of
your limiting reactant fully reacts.
It can be calculated by following these steps:
Calculate moles of all reactants
Determine limiting reactant
Use mole ratios to calculate moles of product expected
Convert moles of product to mass/volume/solution etc.

All this does is string together all the calculations you have already met.
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Some problems

Have a go at the problems found here.
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Key Points
Calculate moles of all reactants
Determine limiting reactant
Use mole ratios to calculate moles of product expected
Convert moles of product to mass/volume/solution etc.
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Lesson 7
Molar Volumes of Gases
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
0.600 mol of aluminium hydroxide is mixed with 0.600
mol of sulfuric acid, and the following reaction occurs:
2 Al(OH)3(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 6 H2O(l)
a.
Determine the limiting reactant.
b.
Calculate the mass of Al (SO ) produced.
2
4 3
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Lesson 7: Equations

Objectives:

Understand that a fixed quantity (in moles) of gas, always
occupies the same volume at room temperature

Perform calculations using the molar volume of a perfect gas

Apply the above concept to design the best possible bottle
rocket.
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The Molar Volume of an Ideal Gas

At standard temperature and pressure (STP):



= 2.24x10-2 m3 mol-1
= 22.4 dm3 mol-1
T = 273K, P = 1.01x105 Pa
At room temperature and pressure (RTP):


Molar Volume of Gas
Molar Volume of Gas
= 2.45x10-2 m3 mol-1
= 24.5 dm3 mol-1
This is true (ish) no matter what the gas!...we will look at
why next lesson
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In Calculations….

What volume of H2 gas is produced when 0.050 mol Mg
reacts with excess acid at S.T.P.?

2 Li(s) + 2 HCl(aq)  2 LiCl(aq) + H2(g)

n(H2) = n(Li) /2 x 1 = 0.050/1 x 1 = 0.025

V(H2) = 0.025 x 22.4 = 0.56 dm3
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More calculations

At RTP, 30 cm3 ethane reacts with 60 cm3 oxygen. Which reactant is
present in excess, how much remains after the reaction and what volume
of CO2 is produced?

2 C2H6(g) + 10 O2(g)  6 H2O(l) + 4 CO2(g)

Limiting reactant:


C2H6: 30/2 = 15
O2: 60/10 = 6


C2H6 remaining:



therefore O2 limiting, ‘6’ will be the number used in all further calculations as there is enough
O2 for ‘6’ of the reaction

V(C2H6 used) = 6 x 2 = 12 cm3
V(C2H6 remaining) = 30 – 12 = 18 cm3
Volume CO2 produced:

V(CO2) = 6 x 4 = 24 cm3
Note: there is no need to convert to moles as they are all in a ratio of moles as
you would divide by 24.0 to get to moles, do your sums and then multiply by
24 .0 to get back to volumes…so why bother?
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Time to practice

What is the minimum volume of H2 gas required to fully reduce 10.0 g copper (II)
oxide to copper (assume RTP…poor assumption but will do for the time being!)?
CuO(s) + H2(g)  Cu(s) + H2O(l)

In a car airbag, sodium azide (NaN3) decomposes explosively to make N2 gas. What
is the minimum mass of sodium azide required to fully inflate a 60.0 dm3 airbag,
assuming RTP?
2 NaN3(s)  2 Na(s) + 3 N2(g)

500 cm3 carbon monoxide reacts with 300 cm3 oxygen to produce carbon dioxide.
What are the final volumes of each of the three gases on completion of the
reaction?
CO(g) + O2(g)  CO2(g)
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Bottle Rockets

Hydrogen reacts explosively with oxygen



Bad if you hate fun!
Excellent if you like to make stuff go bang or
whoosh!
You will need to design and build rockets (from
standard 500 ml drinks bottles). The best will
win an awesome prize.You need to consider:





A suitable reaction to generate hydrogen
The stoichiometry of the reaction
The oxygen content of the air
Suitable ways to decide the winner
Making it look cool (if you finish early!)
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Key Points

The volume of gas depends on the temperature, pressure
and number of moles, NOT THE TYPE OF GAS

Molar Volume at STP = 22.4 dm3

Molar Volume at RTP = 24.5 dm3
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Lesson 8
Ideal Gases
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
What volume of sulfur trioxide, in cm3, can be
prepared using 40 cm3 sulfur dioxide and 20 cm3
oxygen gas by the following reaction? Assume all
volumes are measured at the same temperature and
pressure.
2SO2(g) + O2(g) → 2SO3(g)
A.
B.
C.
D.
20
40
60
80
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Lesson 8: Ideal Gas Equation

Objectives:

Understand the ideal gas equation

Complete a circus of short experiments to explore the ideal
gas equation

Perform calculations using the ideal gas equation
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Molar Volume of a Perfect Gas

We learnt about the molar volume of gases last lesson….how can
they be the same?

The distance between particles is much bigger than the size of the
particles….so particle size makes very little difference:
10 units
10 units

In reality, the relative distance between the molecules is much much
greater than this.
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The Ideal Gas Equation

The volume a gas takes up is determined by:




Pressure
Temperature
Moles of gas
This combines to form the ideal gas equation
PV = nRT

Where:





P = pressure in Pa
V = volume in m3
n = moles of gas
R = gas constant, 8.31 J K-1 mol-1, this appears in many places in chem
T = temperature in K
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Ideal Gas Assumptions

Particles occupy no volume

Particles have zero intermolecular forces

These are not always valid, particularly at:


Low temperature
High pressure
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Study the equation to make some
predictions:

How would does temperature change if you decrease
pressure at fixed volume?

How would volume change if you heat something at fixed
pressure?

How would pressure change if you decrease the volume
at fixed temperature?
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Exploring Ideal Gases

The best way to explore the behaviour of gases is to have
a little play.

Complete these four short experiments
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Some calculations:

1.048 g of unknown gas A, occupies 846 cm3. at 500K
What is it’s molecular mass?
PV

 n( A) 
RT

846
10001000  0.02056
8.31 500
101,000
Mm(A) = mass / moles = 1.048 / 0.02056 = 60.0 g/mol
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More calculations

The volume of an ideal gas at 54.0 °C is increased from 3.00 dm3 to 6.00
dm3. At what temperature, in °C, will the gas have the original pressure?

Use a modified version of the ideal gas equation:
PV
T
1
1

1

PV
T
2
2
2
Since original and final pressure should be the same, we can remove this
from the equation as they cancel out:
54.0 converted to
Kelvin by adding 273
V V
T T
1
2
1
2
3.00 / 327.0 = 6.00 / T2
T2 = (6.00 x 327.0 / 3.00) = 653 K
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Time to practice

Complete the questions found here
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Key Points

The Ideal Gas equation:
PV = nRT

Also:
PV
T
1
1

1

PV
T
2
2
2
Provided that:


Molecules have zero volume
Molecules experience no attraction to each other
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Lesson 9
Test
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Good Luck

You have 80 minutes!
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Lesson 10
Test Debrief
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Personal Reflection

Spend 15 minutes looking through your test:

Make a list of the things you did well

Use your notes and text book to make corrections to
anything you struggled with.
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Group Reflection

Spend 10 minutes working with your classmates:

Help classmates them with corrections they were unable to do
alone

Ask classmates for support on questions you were unable to
correct
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Go Through The Paper

Stop me when I reach a question you still have difficulty
with.
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Targeted Lesson

PREPARE AFTER MARKING THE TEST

SHORT LESSON ON SPECIFIC AREAS OF DIFFICULTY
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