Stoichiometry revised

Report
Introduction to
Mr. Shumway
Chemistry 1

"Stoichiometry" is derived from the Greek
words
 στοιχεῖον (stoicheion, meaning element]) and
 μέτρον (metron, meaning measure.

That’s why it’s a funny looking word

Nevertheless,
 Stoichiometry is the science that deals with
measuring the various elements in chemical
reactions

Write down all that you can about this
reaction below
N2(g) + 3H2(g)  2 NH3(g)

What do you look at when looking at a
chemical equations? There is a lot of
information in a chemical equation…
 Particles
 Moles
 Mass
 Volume
N2(g) + 3H2(g)  2 NH3(g)
Particles
 How many particles are there for each reactant?
 How many particles are there in the product?
There is 1 molecule of Nitrogen reacting with 3
molecules of hydrogen gas
Producing 2 molecules of ammonia
 What is the mole ratio for this reaction?
The mole ratio for this reaction is 1:3:2
N2(g) + 3H2(g)  2 NH3(g)
N
N
N
H
N
H
N2(g) + 3H2(g)  2 NH3(g)
Moles
 How many moles are there for each reactant?
 How many moles are there in the product?
There is 1 mol of Nitrogen reacting with 3 mol
of hydrogen gas
Producing 2 mol of ammonia
 What is the mole ratio for this reaction?
The mole ratio for this reaction is 1:3:2
N2(g) + 3H2(g)  2 NH3(g)
Mass – Law of conservation of mass
 What is the mass of each reactant?
 What is the mass of each product?
 Ex
▪ 1 mol of N2 = 28.0 grams of N2
▪ 1 mol of H2 = 2.0 grams of H2 , therefore, 3 mol of H2 = 6.0
grams of H2
▪ 1 mol of NH3= 17 grams, therefore, 2 mol of NH3 = 34.0 grams
▪ The mass of the reactants = the mass of the products
28.0 + 6.0 = 34.0
N2(g) + 3H2(g)  2 NH3(g)
Volume
 1 mol of a gas at STP = 22.4L
 Therefore, what are the volumes of each gas from
the reactants?
 What is the volume of the product gas that is
formed?
 Ex
▪ 22.4 L of N2 reacts with 67.2L of H2 (3mol x 22.4L)
▪ Forming 44.8L of ammonia
Let’s Look at this problem
2H2S(g) + 3O2(g)  2SO2(g) + 2H2O(g)

Balance the following equation and write
what down all the information that you can
 Number of particles
 Number of moles of reactants and products
 Mass of reactants and products
 Volumes of gases at STP (if applicable)



C2H2(g) + O2(g)  CO2(g) + H2O(g)
Na(s) + H2O(l)  NaOH(aq) + H2(g)
CO(g) +O2(g)  CO2(g)

What do the coefficients tell you in a
chemical reaction?
 What can the coefficients tell you?

What do the coefficients tell you in a
chemical reaction?
 The coefficients tell you the number of
molecules/atoms there are in a chemical reaction.
 What can the coefficients tell you?
▪ They can tell you the mole ratios between each of the
reactants and products
Mole to Mole Ratios
Mr. Shumway
Chemistry 1
We can use the information from the chemical
equation and use them to help us calculate
useful information
 Mole – Mole Calculations
 Mass – Mass Calculations
 Mass-mole and mole mass conversions

Mole – Mole Calculations
 Knowing the amount of moles of reactants, we
can calculate the number of moles of product
▪ Using the mole ratios of the chemical equation

Mole – Mole Calculations
How many moles of ammonia are produced
when 0.60 mol of nitrogen reacts with
hydrogen?
N2(g) + 3H2(g)  2NH3(g)
This is what we
want
This is what we
have
How many moles of ammonia are produced
when 0.60 mol of nitrogen reacts with
hydrogen?
1N2(g) + 3H2(g)  2NH3(g)
What is the ratio/relationship
between these molecules?
How many moles of ammonia are produced
when 0.60 mol of nitrogen reacts with
hydrogen?
N2(g) + 3H2(g)  2NH3(g)
0.60 mol of Nitrogen
X
2 mol of NH3(g)
= 1.2 mol NH3
1 mol of N2(g)
Write down what you
have first
Because we have the number of moles, we need
a conversion factor that has mol of Nitrogen in it
“A mole ratio”
Let’s look at this problem on the board

Calculate the number of moles of reactants
required to make 7.24 mol of Ammonia
N2(g) + 3H2(g)  2NH3(g)
The formation of aluminum oxide from its constituent
elements is represented by this equation
4 Al  3O2 2Al2O3
1.
How many moles of aluminum are needed to form
2.3 mol of Al2O3?
How many moles of oxygen are required to react
completely with 0.84 mol of Al?
2.
3.
Calculate the number of moles of Al2O3 formed
when 17.2 mol of O2 reacts with aluminum.

Design a step-by-step method that will help
you make conversions between moles to
moles in a chemical reaction.
Mole to Mass Calculations
Mr. Shumway
Chemistry 1

Review
 You can calculate the number of moles of product
through the mole ratios of the chemical equation.
▪ Following through with step by step dimensional
analysis

If we can calculate the number of moles, we
can also calculate the mass of the product
and/or the reactants.
A + X  AX
Steps to Follow
1.
2.
3.
It is important that if you are given grams to
first convert to moles because you can use
the mole ratio from the chemical equation
From moles of X to moles of AX (using the
mole ratio)
Convert moles of AX to grams of AX

Calculate the number of grams of NH3
produced by the reaction of 5.40g of
hydrogen with nitrogen
N2(g) + 3H2(g)  2NH3(g)

Step 1: calculate the number of moles by the
mass given
N2(g) + 3H2(g)  2NH3(g)
5.40 g of H2 gas
X
1 mol of H2(g)
= 2.7 mol H2
2.0 g of H2(g)
Write down what you
have first
Because we have the grams of Hydrogen, we
need a conversion factor that has grams/mol
with the proper molecules. Using the molar mass
of the molecule.

Step 2: Convert from moles to moles
N2(g) + 3H2(g)  2NH3(g)
2.70 mol of H2 gas
X
2 mol of NH3(g)
= 1.8 mol NH3(g)
3 mol of H2(g)
Write down what you got
from the last problem
Because we have the moles of Hydrogen, we
need a conversion factor that will give us the
number of moles of ammonia gas, using the
mole ratio in the chemical reaction.
Essentially, we’re converting from moles given to moles
desired

Step 3: Convert from moles to grams
N2(g) + 3H2(g)  2NH3(g)
1.80 mol of NH3(g)
X
17.0 g of NH3(g)
= 30.6 g NH3(g)
1 mol of NH3(g)
Write down what you got
from the last problem
Because we have the moles of ammonia, we
need a conversion factor that will give us the
grams of ammonia gas, using the molar mass of
the molecule.
Step 1
• Grams
of “A”
• Volume
of Gas
“A”
Step 2
• Moles
of “A”
Step 3
• Moles
of “B”
Answer
• Grams
of “B”
• Volume
of Gas
“B”
The combustion of acetylene gas is
represented
by this equation.
2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(g)
1.
2.
How many grams of oxygen are required to
burn 13.0 g of C2H2?
How many grams of CO2 and grams of H2O
are produced when 13.0g of C2H2 reacts with
the oxygen required to burn 13.0g of C2H2?
Acetylene gas, C2H2, is produced by adding
water to calcium carbide, CaC2
CaC2(s) + 2H2O(l)  C2H2 + Ca(OH)2(aq)
1.
2.
How many grams of acetylene are produced
by adding water to 5.00g of CaC2?
How many moles of CaC2 are needed to
react completely with 98.0 g of H2O?
Step 1 Alternate
Step 1
Step 2
• Atoms/particles
of “A”
• Grams of “A”
• Volume of Gas “A”
• Moles of “A”
Step 3
Answer
Alternate
• Moles of “B”
• Grams of “B”
• Volume of Gas “B”
• Atoms/particles
of “B”

Construct a flow chart/map of how to solve
the problems you’ve encountered thus far.

Get into your table groups.

Each group will have a different problem to
work on, you will have 20 min to solve the
problem. Everyone in the group MUST
KNOW HOW TO SOLVE THE PROBLEM
One random person will be chosen to come
to the board and do the problem and explain
their work, this will determine the groups
grade.


What is a limiting reagent that you described
in your mini lab?
Limiting Reagent
Mr. Shumway
Chemistry 1

Limiting reagent
 Limits or determines the amount of product that
can be formed in a reaction

Excess reagent
 Is more than enough to react with a limiting
reagent
Ex. Problem
Sodium Chloride is prepared by the reaction of sodium
metal with chlorine gas. What will occur when 6.70 mol
of Na reacts with 3.20 mol of Cl2?
2Na(s)  Cl2(g ) 2NaCl(s)

What is the limiting reagent?
How many moles of NaCl are produced?
How much excess reagent remains unreacted?

Step 1- What do we know?
 We know/have
▪ 6.70 mol of Na
▪ 3.20 mol of Cl2

Step 2 – What do we want to know?
 We want to calculate three things
▪ What is the limiting reagent?
▪ How many moles of NaCl are produced
▪ How much excess reagent remains unreactedd

Step 3 – Write down a BALANCED chemical
reaction
2Na(s)  Cl2(g ) 2NaCl(s)

Step 4- convert what you know into moles if
not already


Step 5 – Look at the mole ratio and calculate
the amount of substance needed for a
complete reaction
 Use this information to determine the limiting
reagent and excess reagent and label them.
Start with
what we
know
2Na(s)  Cl2(g ) 2NaCl(s)
6.70 mol of Na x

Required amount of
Chlorine
1 mol of Cl2
= 3.35 mol of Cl2
2 mol of Na
Mole Ratio
Knowing that we have
6.70 mol of Na and
knowing the mole ratio
between sodium and
chlorine, we can
calculate the minimum
amount of chlorine we
need for this reaction to
take place
Using the mole ratio, we
can see that 3.35 mol of
Cl2 is needed to react
with 6.70 mol of Na
Because we do not have
enough Cl2,
Cl2 is the limiting
reagent.
How much Chlorine did
the problem say we
had?
Can someone explain
why?
3.20 mol of Cl2!
2Na(s)  Cl2(g ) 2NaCl(s)
Because Cl2 is the limiting
reagent, the amount of
product is determined by the
amount of Cl2 we have.

Therefore, the number of
moles of Cl2 is going to be
used to calculate the
maximum amount of
product that can be formed.
Mole Ratio
3.20 m olCl2 
2 m olNaCl
1 m olCl2
 6.40 m olNaCl
Total amount
of product
formed
HOW MUCH EXCESS DO
WE HAVE? ( Na?)
2Na(s)  Cl2(g ) 2NaCl(s)
The amount of excess reagent
remaining is the difference
between the given amount (in
the beginning of the problem,
a value of 6.70 mol of Na) and
the amount of sodium needed
to react with the limiting 
reagent.

Therefore, we need to now
calculate how much Na
would be used up in the
reaction with Cl2
Mole Ratio
3.20 molCl2 
2 molNa
1 molCl2
 6.40 molNa
Amount of Na
used up in the
reaction
6.70 mol Na – 6.40 mol Na =
0.30 mol Na in excess
Copper reacts with sulfur to form copper (I)
sulfide.
2Cu(s)  S(s) Cu2 S(s)
What is the maximum number of grams of Cu2S
that can be formed when 1.87 mol of Cu
reacts with 2.19 mol of S?

Hydrogen gas can be produced in the
laboratory by the reaction of magnesium
metal with hydrochloric acid.
Mg(s)  HCl(aq ) MgCl2(aq )  H2(g )
How many grams of hydrogen can be produced
when 1.45 mol of HCl is added to 2.31mol of
Mg?
Assuming STP, what is the volume of this
hydrogen?
Write down a systematic approach to dealing
with Limiting reagent type problems. How
would you solve this type of problem, step-by
step?


What is a limiting reagent? How can you
know which reactant is the limiting reagent?
What is an excess reagents? How can you
know which reactant is the excess reagent?
Percent Yield
Mr. Shumway
Chemistry 1
Theoretical Yield
 When an equation is used to calculate the amount
of product that will form during a reaction, the
value is theoretical.
Actual Yield
 The amount of product that forms when the
reaction is carried out in the laboratory
Percent Yield
 Is the ratio of the actual yield to the theoretical
yield. It measure the efficiency of the reaction
PercentYield 
actualyield
theoreticalyield
100%
What is the percent yield of this reaction if
24.8g of CaCO3 is heated to give 13.1g of CaO?
CaCO3(s) CaO(s)  CO2(g )
Looking at this problem,
let’s identify the
important information
This problem tells us the
actual yield, 13.1g of
CaO

What do we want?
We have the actual
yield, now we need the
theoretical.
And we have all we need
to solve for the
theoretical yield.
CaCO3(s) CaO(s)  CO2(g )

Given
24.8 g CaCO3

Converting to moles Mole ratio, looking
using molar mass of at the coefficients
CaCO3
1 m olCaO
1 m olCaCO3


1 m olCaCO3
100.1g CaCO3
This is our
theoretical
yield.

Now we can plug
this into our
equation

Converting to grams
using molar mass of
CaO
56.1 g CaO

1 m olCaO
PercentYield 
actualyield
theoreticalyield
PercentYield 

13.1g CaO
13.9 CaO
PercentYield  94.2%
100%
 100%
What is the percent yield if 3.74g of copper is
produced when 1.87g of aluminum is reacted
with an excess of copper (II) sulfate?
2Al(s)  3CuSO4(aq ) Al2 (SO4 )3(aq )  3Cu(s)
Energy Changes in Chemical Reactions
Mr. Shumway
Chemistry 1
Exothermic Reactions
 Releases energy in the form of heat in a chemical
reaction
Endothermic Reactions
 Energy is absorbed in a chemical reaction
Thermochemical Equation
 An equation that includes the amount of heat
produced or absorbed by a chemical reaction.
Exothermic Reaction
“Releases Energy”
Or gives off energy
Where would the energy be in the
chemical reaction?
The products or reactants?
C(s) O2(g ) CO2(g )  393.4KJ
The Energy would be in the products

Endothermic Reaction
“Absorbs Energy”
Or requires energy
Where would the energy be in the
chemical reaction?
The products or reactants?
CaCO3(s)  176KJ CaO(s)  CO2(g)
This equation is called a thermochemical
equation
The Energy would be in the
reactants
We can use the thermochemical equation to
help us calculate the energy required for a
chemical equation.
Think of this as another conversion factor to
add
Using the equation just given, calculate the
energy required to decompose 5.20 mol of
CaCO3 (in Kilojoules)
1CaCO3(s)  176KJ CaO(s)  CO2(g)
5.20 mol CaCO3


176 KJ
1 m olCaCO3
= 915 KJ
Enthalpy
Is the amount of heat that a substance has at
a given temperature and pressure.
Is symbolized by “H”
If an exothermic reaction is taking place then the
reaction has a “-ΔH”
If an endothermic reaction is taking place then
the reaction has a “+ΔH”
Standard Heat of Formation
is where the “ΔH” for a reaction in which one
mole of a compound is formed from its
elements of that compound.
The air pollutant sulfur trioxide reacts with
water in the atmosphere to produce sulfuric
acid and heat.
SO3(g )  H2O(l ) H2SO4(aq ) 129.6 KJ
How much heat is released when 583 g of SO3(g)
reacts with water?
Carbon dioxide can be decomposed into carbon
monoxide and oxygen by the absorption of
heat.
2CO2(g )  43.9 KJ 2CO(g )  O2(g)
How many molecules of CO2(g) can be
 decomposed by the addition of 22.2KJ of
heat energy?

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