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StoichiometryChapter 9
Stoichiometry – Chapter 9Stoichiometry is the study of the quantities of materials
consumed and produced in chemical reactions.What do you need to know to be able to complete this
chapter? Chapter 7 – Chemical Quantities – moles/representative
particles/molar mass/molar volume/MOLE ROADMAPChapter 8 – Balancing Chemical EquationsChapter 3&4 – Dimensional Analysis
S’moresWhen you have a beach BBQ, you make s’mores, right?When I make s’mores, here’s what my chemical reaction
looks like:
2 graham crackers + 1 marshmallow + 1 piece chocolate → 1 s’more
Let’s say I wanted to make 8 s’moresfor my friends. How much of each Ingredient would I need? That’s stoichiometry!
S’mores
That’s a conversion factor for making s’mores
Why stoichiometry?Let’s say you want to relate the amount of product made
in a chemical reaction to the amount of reactants you need to start off with.
N2 + 3 H2 → 2 NH3
So let’s go back to previous example:N2 + 3 H2 → 2 NH3 + →
2 atoms N + 6 atoms H → 2 atoms N and 6 atoms H1 molecule N2 + 3 molecules H2 → 2 molecules NH3
10 molecule N2 + 30 molecules H2 → 20 molecules NH3
1(6.02E23) molecules N2 + 3(6.02E23) molecules H2 → 2(6.02E23) molecules NH3
1 mol N2 + 3 mol H2 → 2 mol NH3 (mol reactants ≠ mol products!)(28.0) g N2 + 3 (2.0 g) H2 → 2 (17.0) g NH3
34 g reactants → 34 g products (conservation of mass/matter)
22.4 L N2 + 3(22.4) L H2 → 2(22.4) L NH3 (L reactants ≠ L products)
Mole-Mole CalculationsN2 (g) + 3 H2 (g) → 2 NH3 (g)What would the six mole/mole ratios be for this
reaction? These are the 6 possible conversion factors associated with this reaction.
Mole/mole ratios give the ratio of how much of two of the chemicals are needed to react together in this reaction.
Mole-Mole CalculationsSo let’s say I started out with 0.4 mol of N2. How much
ammonia (how many moles) could I produce using that reaction?
If I did make 0.8 mol of NH3, how much H2 would I have to supply to make that happen?
Let’s try another exampleFirst we have to balance the following equation ___ C5H12 + ____O2 → ____CO2 + ____H2O
Step 1: C 1 C5H12 + ____O2 → 5 CO2 + ____H2O
Step 2: H 1 C5H12 + ____O2 → 5 CO2 + 6 H2O
Step 3: O 1 C5H12 + 8 O2 → 5 CO2 + 6 H2O
Total of 5 C, 12 H and 16 O to balance it.
Mole-Mole CalculationsWe can now relate moles of product to the moles of
reactant (or any species involved in the reaction, really) in mathematical equations.
Mole ratio: A conversion factor derived from the coefficients of a balanced equation.You have to start with a balanced equation.Use a mole ratio when you want to relate one chemical’s
quantity in the reaction to another chemical’s quantity.
Mole Ratios1 C5H12 + 8 O2 → 5 CO2 + 6 H2OLet’s try a few mole ratios using the balanced equation
from a few pages ago.How many moles of water are produced when 4 moles of
oxygen are reacted? Start with writing the given 4 moles oxygen on the left…Then write what is wanted on the right… moles of waterThen use an appropriate mole-mole conversion factor
Mole ratios1 C5H12 + 8 O2 → 5 CO2 + 6 H2OHow many moles of pentane are needed to produce 12
moles of carbon dioxide?Start by writing the given on the left (12 moles CO2)Next write the thing you want to find on the right (moles
of pentane)Then use an appropriate conversion factor
You try it on whiteboardsGiven the equation 4 Al (s) + 3 O2 (g) → 2 Al2O3 (s)Write the six possible mole ratios on the white board.
How many moles of Al are needed to form 3.7 moles of Al2O3 ? Show the calculation on the white board.
How many moles of O2 are required to react with 14.8 moles of Al?
How many moles of Al2O3 are formed when 0.78 moles of O2 react with excess Al?
Now recall the mole roadmap
Stoichiometry Concept MapAll roads lead to the balanced equation and mole ratios!
Problems not given in moles:Example: How many moles of carbon dioxide will be
produced when 10 grams of pentane react with excess oxygen gas?
Note how the question is not moles to moles - it is moles and grams, so we need to convert.
Strategy:1. Convert the given quantity in the problem to moles. If it
is already given in moles, skip this step.2. Use the mole ratio. Hint: you will never skip this step.3. Convert the value to the unit asked for in the problem.
Let’s try an example1 C5H12 + 8 O2 → 5 CO2 + 6 H2OHow many moles of carbon dioxide will be produced when
20.5 grams of pentane reacts with excess oxygen?What goes on the left side of the paper? What goes on the right side of the paper?What is the molar mass of pentane? 5(12.0) + 12(1.0) = 72 g/mole pentane
molar mass pentane mole/mole conversion factor
You try this one on your whiteboard1 C5H12 + 8 O2 → 5 CO2 + 6 H2OHow many grams of oxygen gas are needed to react if
12.8 moles of water are formed?What do you write on the left side of the board?
What do you write on the right side of the board? What do you need the molar mass of?
You should get an answer of about 546 g of O2 – did you?
Mass-Mass conversionsNow what if both the “given” and the desired answer are
in grams rather than moles?How many grams of water will be formed when 552.3
grams of pentane are reacted in the presence of oxygen?1 C5H12 + 8 O2 → 5 CO2 + 6 H2O
=
882.5 g H2OHere, because both given and answer were in grams, you
had to use two molar mass conversion factors, plus the mole ratio conversion factor (that you always use).
Representative particle practice Using the reaction: 2 KClO3 → 2 KCl + 3 O2
How many molecules of O2 are produced by the decomposition of 6.54 grams of KClO3 ?
First think about what goes on far left and far right.Then figure out what mole-mole conversion factor to use.Then you need one molar mass KClO3 =
39.1+35.4+3(16.0)=122.5Then you need Avogadro’s number conversion factor
Volume of a gas practiceGiven the reaction: 2 CO + 1 O2 → 2 CO2
How many liters of O2 are required to burn 3.86 liters of CO?Recall the conversion factor for gases at STP that
22.4 liters = 1 mole.What do you start with on the left? On the right?
Do you see that the 2nd and 4th term look like they cancel each other out? Careful, your units won’t work out if you skip them and you won’t get credit on the test if you skip them.
Try one on your whiteboardGiven the reaction: 2 H2O → 2 H2 + O2
How many molecules of oxygen are produced when a sample of 29.2 g of water is decomposed by electrolysis according to the equation above?
Last one on your whiteboardGiven the reaction: 2 SO2 (g) + O2 (g) → SO3 (g)First of all, is that balanced?Once you have it balanced, then assuming STP (standard
temperature and pressure), how many liters of oxygen are needed to produce 19.8 L SO3 ?
What goes on the left?What goes on the right? Are you going to need any molar masses?
Section 9.3 Limiting ReagentsLet’s go back to our discussion of s’mores.Let’s say I want to make 40 s’mores. Each s’more is made up of
2 graham crackers1 marshmallow1 square of chocolate
If I have 70 graham crackers, 42 marshmallows and 45 squares of chocolate, how many s’mores can I make?
The limiting factor is the graham crackers. I can only make 35 s’mores from the graham crackers I have. That is the limiting reagent. I have enough marshmallows and chocolate to make 40, so they are NOT limiting reagents.
Limiting reagentsLet’s say I start out with 2 moles of N2 and 3 moles of H2 in a container and I
want to make ammonia.
N2 + 3 H2 → 2 NH3 This is the balanced reaction.Which reactant will be the limiting reagent? Let’s look at the hydrogen first:3 moles of hydrogen require 1 mole of nitrogen to make 2 moles of ammonia
gas. Do I have enough nitrogen to do that?Yes, I have 2 moles of nitrogen and I only need 1 mole (there is excess
nitrogen). Now let’s look at the nitrogen instead:2 moles of nitrogen would mix with 6 moles of hydrogen to give 4 moles of
NH3 gas (refer to the balanced equation to see that).Do I have enough hydrogen to do that?No, I need 6 moles of hydrogen but I only have 3 moles of it. So hydrogen is
the limiting reagent.
Limiting reagents Looking at it graphically, which reactant was limited?(which reactant had excess, and which got all used up?)Before reaction After reactionN2 + 3 H2 → 2 NH3 N2 + 3 H2 → 2 NH3
Limiting reagents example 9-92 Cu + S → Cu2SWhat is the limiting reagent when 80.0 g of Cu reacts
with 25.0 g S ?Need to know how many moles that is for Cu and for S.
If we compare the amount of S needed (0.630 mole) to the amount of S we have (0.779 mole), we have enough S so it is not the limiting reagent, Cu is the limiting reagent.
This is how much of each reactant we HAVE
Section 9.3: Percent yieldIf you do an experiment and mix two reactants and
expect a certain amount of products, do you always get that much?
If there is experimental error (and there will be), then you will never get exactly that much.
The percent yield for your experiment is :
Percent yield – example 9-10Calcium carbonate is decomposed by heating:CaCO3 → CaO(s) + CO2
Find the theoretical yield and the percent yield if 13.1 g of CaO is produced.
Given: the starting mass of calcium carbonate is 24.8 g.
Molar mass of CaCO3 is 100.1 g/molMolar mass of CaO is 56.1 g/mol Mole ratio of CaO to CaCO3 is 1:1