chapter 12 stoichiometry the study of the quantitative, or measurable, relationships that exist in...

Chapter 12Chapter 12

StoichiometryStoichiometry

The study of the quantitative, or measurable, relationships that exist in chemical formulas and chemical reactions.

StoichiometryStoichiometry

2 H2 H22 + O + O22 2 H 2 H22OO

Interpreting Balanced Chemical EquationsInterpreting Balanced Chemical Equations

• Based on the mole ratio

Mole RatioMole RatioMole RatioMole Ratio

• indicated by coefficients in a balanced equation

Molar ratio of H2 to H2O is 2:2

(Simplify 1:1)

Molar ratio of O2 to H2O is 1:2

2 H2 H22 + O + O22 2 H 2 H22OO

2 moles of hydrogen react with

1 mole of oxygen to produce 2

moles of water.

2 H2 H22 + O + O22 2 H 2 H22OO

In terms of molesIn terms of moles

4 grams of hydrogen react with

32 grams of oxygen to produce

36 grams of water.

2 H2 H22 + O + O22 2 H 2 H22OO

In terms of massIn terms of mass

2 hydrogen molecules react with 1 oxygen

molecule to produce 2 water molecules.

• Notice that the number of molecules is

NOT the same on each side of the arrow.

2 H2 H22 + O + O22 2 H 2 H22OO

In terms of moleculesIn terms of molecules

4 atoms of hydrogen react with 2

atoms of oxygen to produce 2

water molecules, which are 4

atoms of hydrogen and 2 atoms

of oxygen.

2 H2 H22 + O + O22 2 H 2 H22OO

In terms of atomsIn terms of atoms

44.8 L of hydrogen gas react with 22.4 L

of oxygen gas to produce 44.8 L of water

vapor.

• Notice that the number of liters of gas is

NOT the same on each side of the arrow.

2 H2 H22 + O + O22 2 H 2 H22OO

In terms of volumesIn terms of volumes

Law of Conservation of Matter & Law of Conservation of Matter & MassMassLaw of Conservation of Matter & Law of Conservation of Matter & MassMass

2 moles H2 react with 1 mole of

O2 to form 2 moles of H2O.

2 H2 H22 + O + O22 2 H 2 H22OO

HH22 OO22 HH22OO2 mol 1 mol 2 mol2 g

1 mol

32 g

1 mol

18 g

1 mol

4 g 32 g 36 g++ ==

VERIFIED!VERIFIED!VERIFIED!VERIFIED!

Lead will react Lead will react with hydrochloric with hydrochloric acid to produce acid to produce lead (II) chloride lead (II) chloride and hydrogen. and hydrogen. How many moles How many moles of hydrochloric of hydrochloric acid are needed acid are needed to completely to completely react with 0.36 react with 0.36 moles of lead?moles of lead?

Practice Problem #1Practice Problem #1

Oh, no … Oh, no … where do I where do I start?start?

Lead will react with hydrochloric acid Lead will react with hydrochloric acid to produce lead chloride and to produce lead chloride and hydrogen. hydrogen.

Write an equation and balance it.Write an equation and balance it.

Pb + HCl Pb + HCl PbCl PbCl22 + H + H2222

How many moles of How many moles of hydrochloric acidhydrochloric acid are needed to completely react with are needed to completely react with 0.36 moles of 0.36 moles of leadlead??

Determine Mole RatioDetermine Mole Ratio

Pb + HCl Pb + HCl PbCl PbCl22 + H + H2222

hydrochloric hydrochloric acidacid

leadlead

Coefficient = 1Coefficient = 1Coefficient = 2Coefficient = 2

““wanted”wanted”““given”given”

How many moles ofHow many moles of hydrochloric acidhydrochloric acid are needed to completely react with are needed to completely react with 0.36 moles of0.36 moles of leadlead??

Set up mole ratio “wanted” to “given”Set up mole ratio “wanted” to “given”

PbPb + + HClHCl PbCl PbCl22 + H + H2222

wantedwanted

leadlead 11

22hydrochloric acidhydrochloric acid

givengiven == ==

0.36 moles of0.36 moles of leadlead

X moles of HClX moles of HCl==

11

22

How many moles ofHow many moles of hydrochloric acidhydrochloric acid are needed to completely react with are needed to completely react with 0.36 moles of0.36 moles of leadlead??

Solve for Solve for “wanted”“wanted”

PbPb + + HClHCl PbCl PbCl22 + H + H2222

0.36 moles of0.36 moles of leadlead

X moles of HClX moles of HCl==

11

22

X = 2 X = 2 (0.36 moles)(0.36 moles)

X = 0.72 moles HClX = 0.72 moles HCl

Umm … It would Umm … It would require 0.72 moles require 0.72 moles of hydrochloric of hydrochloric acid. Is that right?acid. Is that right?

That’s Correct!That’s Correct!

Hem, hem –Hem, hem –you have 9 more you have 9 more problems … problems … ready, go.ready, go.

How to solve, given an amount of one substance

to another substance.

How to solve, given an amount of one substance

to another substance.

““Stoich” ProblemsStoich” Problems

Stoichiometry StepsStoichiometry StepsStoichiometry StepsStoichiometry Steps

1. Write a balanced equation.2. Identify “wanted” & “given”.3. Convert given information to moles.4. Determine Mole Ratio.

(Moles of “wanted” : Moles of “given”)

5. Calculate Moles of “wanted”.6. Convert to required units.

KEY step in all stoichiometry problems!

4. Determine Mole ratio (moles of “wanted” to moles of “given”)

What mass of bromine is produced when fluorine reacts with 1.72 g of potassium bromide?

Mass-Mass ProblemsMass-Mass Problems

Help … I really need to know this right now!

What mass of bromine is produced when fluorine reacts with 1.72 g of potassium bromide?

1. Write a balanced equation.1. Write a balanced equation.

Help … I really need to know this right now!

FF22 + 2KBr + 2KBr 2KF + Br 2KF + Br22

What mass of bromine is produced when fluorine reacts with 1.72 g of potassium bromide?

2. Identify 2. Identify “wanted” “wanted” and and “given”“given”

Help … I really need to know this right now!

FF22 + + 2KBr2KBr 2KF + 2KF + BrBr22

““wanted”wanted”““given”given”

1.72 g of potassium bromide

3. Convert 3. Convert “given” “given” information to molesinformation to moles

Help … I really need to know this right now!

Molar Mass of KBr = 39 + 80 = 119 g/mol

1.72 g

119 g

1 mol= 0.01445 mol KBr

4. Determine Mole Ratio4. Determine Mole Ratio

Help … I really need to know this right now!

FF22 + + 2KBr2KBr 2KF + 2KF + BrBr22

““wanted”wanted”““given”given”

Coefficients tell Mole Ratio Coefficients tell Mole Ratio ““wanted” wanted” toto “given”“given”

11 :: 22

5. Calculate Moles of 5. Calculate Moles of “wanted”“wanted”

Help … I really need to know this right now!

FF22 + + 2KBr2KBr 2KF + 2KF + BrBr22

Moles KBrMoles KBr

Moles BrMoles Br22 11==

22

0.01445 mol KBr==

XX 11

22

2 2 X = X = 0.01445 mol X = X = 0.007225 mol

6. Convert to required units6. Convert to required units

X = X = 0.007225 mol BrBr22

0.007225 mol

Molar Mass BrMolar Mass Br22 80 + 80 = 160 g/mol80 + 80 = 160 g/mol

160 g160 g

1 mol1 mol== 1.16 g Br1.16 g Br22

StoichiometryStoichiometry

Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants

Available IngredientsAvailable Ingredients• 4 slices of bread• 1 jar of peanut butter• 1/2 jar of jelly

Limiting ReactantLimiting Reactant• bread

Excess ReactantsExcess Reactants• peanut butter and jelly

Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants

Available IngredientsAvailable Ingredients• Copper Wire• 0.5 g AgNO3

Limiting ReactantLimiting Reactant• 0.5 grams AgNO3

Excess ReactantsExcess Reactants• Copper Wire

Limiting ReactantLimiting ReactantLimiting ReactantLimiting Reactant

The reactant that limits the The reactant that limits the amount of product that can be amount of product that can be formed.formed.

When quantities of reactants are When quantities of reactants are available in the exact ratio available in the exact ratio described by the balanced described by the balanced equation, they are said to be in equation, they are said to be in Stoichiometric proportions.Stoichiometric proportions.

Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants

Limiting ReactantLimiting Reactant• used up in a reaction• determines the amount of all

products formed

Excess ReactantExcess Reactant• added to ensure that the other

reactant is completely used up• usually cheaper & easier to recycle

Solving Problems – Limiting Solving Problems – Limiting ReactantsReactantsSolving Problems – Limiting Solving Problems – Limiting ReactantsReactants

1. Write a balanced equation.

2. For each reactant, calculate the

amount of product formed.

3. The reactant that produces the smaller amount of product is the limiting reactant.

Very similar to mass-mass problems!

Step 1: Write a balanced Step 1: Write a balanced equation.equation.Step 1: Write a balanced Step 1: Write a balanced equation.equation.

Identify the limiting reactant when 1.22 g of oxygen reacts with 1.05 g of hydrogen to produce water.

O2 + 2H2 2 H2O

Step 2:Step 2:Step 2:Step 2:

Identify the limiting reactant when 1.22 g of oxygen reacts with 1.05 g of hydrogen to produce water.

O2 + 2H2 2 H2O

For each reactant,

calculate the amount of

product formed.

Step 2: Step 2: Step 2: Step 2:

1.22 g oxygen

O2 + 2H2 2 H2O

wanted

given

wanted

=X

0.038 mol

2

1

given

X = 0.076 mol H2O

32 g

1 mole= 0.038 mol O2

Step 2: Step 2: Step 2: Step 2:

1.05 g H2

O2 + 2H2 2 H2O

wanted

given

wanted

=X

0.525 mol

2

2

given

X = 0.525 mol H2O

2 g

1 mole= 0.525 mol H2

Step 3:Step 3:Step 3:Step 3:

1.22 g of O2 would produce 0.0763 mol H2O

The one that produces the smallest amount is your

limiting reactant.

1.05 g of H2 would produce .525 mol H2O

Oxygen is your limiting reactant!

Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants

Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride.

Step 1: Write a balanced Step 1: Write a balanced equation.equation.Step 1: Write a balanced Step 1: Write a balanced equation.equation.

Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride.

2Na + Cl2 2NaCl

Step 2:Step 2:Step 2:Step 2:

Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride.

2Na + Cl2 2NaCl

For each reactant,

calculate the amount of

product formed.

Step 2: Step 2: Step 2: Step 2:

1.7 g Na

2Na + Cl2 2NaCl

wanted

given

wanted

=X

0.0739 mol

2

2

given

X = 0.0739 mol NaCl

23 g

1 mole = 0.0739 mol Na

Step 2: Step 2: Step 2: Step 2:

2.6 L Cl2

2Na + Cl2 2NaCl

wanted

given

wanted

=X

0.116 mol

2

1

given

X = 0.232 mol NaCl

22.4 L

1 mole = 0.116 mol Cl2

Step 3:Step 3:Step 3:Step 3:

1.7 g Na would produce 0.0739 mol NaCl

The one that produces the smallest amount is your

limiting reactant.

2.6 L Cl2 would produce 0.232 mol NaCl

Sodium is your limiting reactant!

Click on the real player file called Sodium_Chlorine_2 to see a demo of this reaction

Percent YieldPercent YieldPercent YieldPercent Yield

100yield ltheoretica

yield actualyield %

calculated on paper

measured in lab

Percent YieldPercent YieldPercent YieldPercent Yield

When 45.8 g of K2CO3 react with

excess HCl, 46.3 g of KCl are formed. Calculate the theoretical yield and % yield of KCl.

K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ? g

actual: 46.3 g

Percent YieldPercent YieldPercent YieldPercent Yield

45.8 gK2CO3

1 molK2CO3

138 gK2CO3

= 49.1g KCl

2 molKCl

1 molK2CO3

74g KCl

1 molKCl

K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ? g

actual: 46.3 g

Theoretical Yield:

Percent YieldPercent YieldPercent YieldPercent Yield

Theoretical Yield = 49.1 g KCl

% Yield =46.3 g

49.1 g 100 = 94.3%

K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g 49.1 g

actual: 46.3 g