Unit 5
Reactions and Stoichiometry
Chemistry
Ch 11
Describing Chemical Reactions
Introduction to Chemical Equations
• ALL chemical reactions follow 2 laws!
• Law of Conservation of mass:
– In a chemical reaction, the total mass of the reactants
and the total mass of the products must be the same.
• Law of Conservation of Atoms:
– In a chemical reaction, atoms can not be created or
destroyed.
• Chemical Reactions are about the arrangement of
Atoms, Not the changing of Atoms
• The only thing about atoms that changes to drive
chemistry is the location of electrons
• Reaction Vocabulary:
•
•
•
•
Reactants: Starting Atoms, Molecules, or Compounds
Products: Ending Atoms, Molecules, or Compounds
Yields: Progress of reaction – Gives “→”
Catalyst: Substance that speeds up chemical reactions
Reactants
Catalyst
Products
– Additional Information
• States of Matter
– Shown as (s), (l), (g), (aq) after formulas
• Catalyst
– Shown above Yield arrow
– If Heat is added show with ∆ or ∆H above Yield arrow
• Example:
– Aqueous hydrogen peroxide will decompose into
gaseous forms of oxygen and water in the
presence of manganese dioxide as a catalyst
H 2O2 ( aq)  O2 ( g )  H 2O( g )
MnO2
Balancing Chemical Equations
Following the Law of conservation of ATOMS
• You must NEVER change the # of Atoms in a formula by
changing the subscript numbers
• To change # of atoms (or molecules or compounds), add a
coefficient
• Coefficients are numbers placed in front of a formula.
• Coefficients change the # of atoms by a multiplying effect
• Example:
– Hydrogen gas combusts with Oxygen gas to form
water.
1. Get Formulas correct
2. Organize REACTANTS vs. PRODUCTS
3. BALANCE
• Balancing Reaction HINTS
– Look for Polyatomic Ions!
– Save elements O2 and H2 for the end.
– If you assume the coefficient of a large particle is 1
and come to a dead end, change it to a 2.
Mg3(PO4)2 +
C2H6 + O2
NaCl 

MgCl2 +
CO2 + H2O
Na3PO4
Reaction Types
1. Combination Reaction:
– 2 or more substances as Reactants combine to form 1
Product
– Also Called “Synthesis”
2. Decomposition Reaction:
– 1 Reactant breaks down into 2 or more Reactants
– Most require energy to take place (take in Energy)
3. Single-Replacement Reaction:
– 1 element replaces another element in a compound
– Reactivity of Metals and Non-Metals is important:
– Use activity series (Page 361).
4. Double-Replacement Reaction:
–
–
–
–
An Exchange of positive ions between 2 Ionic Compounds
Both Reactants are Ionic Compounds
Often takes place in solution with water.
Sometimes a Product is a solid that will not dissolve in water
and forms a PRECIPITATE
5. Combustion Reactions
– A Reactant combines with Oxygen (other Reactant)
– Usually produce energy (give off energy)
– Usually involves a Molecular Compounds with (C,H, & or O) and
Oxygen
– Often have Water and Carbon dioxide as Products
• “Burning Hydrocarbons”
• Wood, Gas, Fuel, Sugars (Remember Cellular Respiration?)
– May also be a Combination (synthesis) Reaction if only one
Product forms
• “Burning Elements”
• Combination:
A + B  AB
• Decomposition:
AB  A + B
• Single Replacement:
AB + C  CB + A
• Double Replacement:
AB + CD  AD + CB
• Combustion:
? + O2  H2O + CO2
• Predicting Reaction Types Examples
1.
2.
3.
4.
5.
A single Reactant: H2O2
A metal (Na) and an Ionic Compound (MgCl2) React:
C2H6 reacts with O2:
Two Ionic compounds react [KCl & Mg(NO3)2]
A single product: H2O
12.1 The Arithmetic of Equations
• Everyday equations
• What’s a good equation for a tricycle made of the following
components?
–
–
–
–
–
Frame
Seat
Wheels
Handbars
Pedals
(F)
(S)
(W)
(H)
(P)
F + S + 3 W + H + 2 P  FSW3HP2
But what if Wheels and Pedals only came in pairs where the
manufacturer of the tricycles get their parts?
F + S +
W 2 + H + P2 
• What’s a good equation for a tricycle made of the following
components?
–
–
–
–
–
Frame
Seat
Wheels
Handbars
Pedals
(F)
(S)
(W)
(H)
(P)
F + S + 3 W + H + 2 P  FSW3HP2
But what if Wheels and Pedals only came in pairs where the
manufacturer of the tricycles get their parts?
2F + 2S + 3W2 + 2H + 2P2  2FSW3HP2
Chemical Equations
• Stoichiometry:
– The calculation of quantities in chemical reactions
– Based on a balanced chemical equation
• What does a balanced chemical reaction REALLY
MEAN?
– Example: Photosynthesis
6 H2O + 6 CO2 → C6H12O6 + 6 O2
• Mole Ratio:
– Ratio of moles of substances in a balanced chemical
reaction.
– Based on Coefficients in front of formula
6 H2O (l) + 6 CO2 (g) → C6H12O6 (s) + 6 O2 (g)
• For every 1 Glucose _Molecule__there needs to be?
___ H2O _______
___ CO2 _______
___ O2 _______
6 H2O (l) + 6 CO2 (g) → C6H12O6 (s) + 6 O2 (g)
• For every 1 Glucose _Molecule__there needs to be?
_6_ H2O Molecules
_6_ CO2 Molecules
_6_ O2 Molecules
6 H2O (l) + 6 CO2 (g) → C6H12O6 (s) + 6 O2 (g)
• For every 1 Glucose __Mole__there needs to be?
___ H2O _______
___ CO2 _______
___ O2 _______
6 H2O (l) + 6 CO2 (g) → C6H12O6 (s) + 6 O2 (g)
• For every 1 Glucose __Mole__there needs to be?
_6_ H2O __ Mole _
_6_ CO2 __ Mole _
_6_ O2 __ Mole _
• 1 mole of Glucose is ____________ g
6 (12.01) =
12 (1.01) =
6 (16.00) =
• So for every _________ g of Glucose
_________ H2O
_________ CO2
_________ O2
• 1 mole of Glucose is ____________ g
6 (12.01) = 72.06 g
12 (1.01) = 12.12 g
6 (16.00) = 96.00 g
• So for every _________ g of Glucose
_________ H2O
_________ CO2
_________ O2
• 1 mole of Glucose is __
6 (12.01) = 72.06 g
12 (1.01) = 12.12 g
6 (16.00) = 96.00 g
180.18 g
• So for every _________ g of Glucose
_________ H2O
_________ CO2
_________ O2
• 1 mole of Glucose is __
6 (12.01) = 72.06 g
12 (1.01) = 12.12 g
6 (16.00) = 96.00 g
180.18 g
• So for every ___ 180.18 g of Glucose
_________ H2O
_________ CO2
_________ O2
• 1 mole of Glucose is __
6 (12.01) = 72.06 g
12 (1.01) = 12.12 g
6 (16.00) = 96.00 g
180.18 g
• So for every ___ 180.18 g of Glucose
_108.12 g_ H2O
_264.06 g_ CO2
_192.00 g_ O2
• So
6 H2O (l) + 6 CO2 (g) → C6H12O6 (s) + 6 O2 (g)
• Sample Problem 12.2 (page 388)
Hydrogen sulfide, which smells like rotten eggs, is
found in volcanic gases. The balanced equation
for the burning of hydrogen sulfide is:
2 H2S (g) + 3 O2 (g)  2 SO2 (g) + 2 H2O (g)
Interpret this equation in terms of:
a. numbers of representative particles and moles.
b. masses of reactants and products.
12.2 Chemical Calculations
Writing and Using Mole Ratios: Page 390
• Mole Ratio:
– Ratio of moles of substances in a balanced
chemical reaction.
– Based on Coefficients in front of formula
2H2O2  2H2O + O2
• What are all the mole ratios of this equation?
2molH 2O2  2molH 2 0
2molH 2O2  1molO2
2molH 2O  1molO2
• If 4.3 moles of Oxygen gas is produced, how
many moles of water will also be produced?
2molH 2 0
4.3molO2
 8.6molH 2O
1molO2
Sample Problem 12.3 (page 391)
How many moles of NH3 are produced when
0.60 mol of nitrogen reacts with hydrogen in
the equation?
N2 (g) + 3 H2 (g)  2 NH3 (g)
Stoichiometry: Mass - Mass Calculations
• Using balanced reactions and mole ratios to predict
measurable quantities in a chemical reaction.
• Example: MASS
– How many grams of glucose (C6H12O6) are
produced if 120 g of CO2 react completely with
enough water in photosynthesis?
• Steps
1. Make sure equation is balanced, balance if need
be
6 H2O + 6 CO2 → C6H12O6 + 6 O2
2. Write down given value
3. Convert given value to moles
4. Convert moles substance to moles other
substance
•
USE Mole Ratio
5. Convert moles to mass
1molCO2 1molC6 H12O6 180.0 gC6 H12O6
120 gCO2
120gCO
44.0 gCO2 6molCO2
1molC6 H12O6
Sample Problem 12.4 (page 393)
Calculate the number of grams of NH3
produced by the reactions of 5.40 g of H2 with
an excess of N2. The balanced equation is:
N2 (g) + 3 H2 (g)  2 NH3 (g)
Other Stoichiometric Calculations
# of
Particles
# of
Particles
# Moles
Substance A
Mass
# Moles
Substance B
Mass
3 Cu(s) + 8 HNO3(aq)  3 Cu(NO3)2(aq) + 4 H2O(l) + 2 NO(g)
• How many moles of water are produced from 2.5 moles copper?
• How many moles of NO are produced from 33.5g of HNO3?
• How many grams of copper are needed to produce 1.25g of water?
• How many grams of NO are produced if 3.20x1023 molecules of
water are produced?
Sample Problem 12.5 (page 395)
How many molecules of oxygen are produced
when 29.2 g of water is decomposed by
electrolysis according to this balanced
equation?
2 H2O (l)  2 H2 (g) + O2 (g)
12.3 Limiting Reagent and Percent Yield
• Limiting Reagent:
– Determines the amount of product that can/will
form
– Reactant that is completely used up in a reaction
• Excess Reagent:
– Reactant that is NOT completely used up in a
reaction
• Limiting Reagent Analogy: HAM SANDWICH
1 slice ham + 1 slice cheese + 1 slice tomato + 2 slices bread  1 ham sandwich
– Actual Ingredients:
•
•
•
•
4 slices of bread
4 slices of tomato
10 slices of ham
1 slice of cheese
• What is the Limiting Reagent?
• How to find limiting reagents
– Need to analyze MOLES
– Find the # of Moles for each reactant based on the
# of moles of the other reactants.
– Compare the # of Moles needed to the actual # of
Moles present.
– If amounts are given in other units (g’s?), covert to
Moles first.
• Example: C2H4 + 3O2  2CO2 + 2 H2O
– If 2.70 mol C2H4 is reacted with 6.30 mol O2:
– Find the limiting reagent
3molO2
2.70molC2 H 4
 8.10molO2
1molC2 H 4
1molC2 H 4
6.30molO2
 2.10molC2 H 4
3molO2
– Calculate the moles of water produced
• Must use LIMITING REAGENT!
2molH 2O
6.30molO2
 4.20molH 2O
3molO2
Another Example
2Al(s) + 3CuSO4(aq) → 3Cu(s) + Al2(SO4)3(aq)
1. Which reactant is the limiting reagent?
2. What mass of copper is produced
3. How much (mass) excess reagent is left over?
1. Which reactant is the limiting reagent?
• Using the actual amount of Al (25.0 g), find
how much CuSO4 would be needed.
1molAl 3molCuSO4 159.6 gCuSO4
25.0 gAl
 222 gCuSO4
27.0 gAl 2molAl
1molCuSO4
• Compare 222 g CuSO4 needed to actual 75.0g
• CuSO4 is the LIMITING REAGENT
2. What mass of copper is produced
• Use the 75.0 g CuSO4 Limiting Reagent
1molCuSO4
3molCu 63.5 gCu
75.0 gCuSO4
 29.8 gCu
159.6 gCuSO4 3molCuSO4 1molCu
• 29.8 g Cu should be formed
3. How much (mass) excess reagent is left over?
• Need to solve for actual amount of excess
reagent that reacted. Use Limiting reagent.
1molCuSO4
2molAl 27.0 gAl
75.0 gCuSO4
 8.46 gAl
159.6 gCuSO4 3molCuSO4 1molAl
• Started with 25.0 g Al and used 8.46 g Al
25.0 gAl  8.46 gAl  16.5 gAl
• 16.5 g Al will remain after the reaction is
complete
Sample Problem 12.8 (page 402)
Copper reacts with sulfur to form copper (I)
sulfide according to the following balanced
equation:
2 Cu (s) + S (s)  Cu2S (s)
What is the limiting reagent when 80.0 g Cu
reacts with 25.0 g S?
Percent yield
• The actual yield is the amount of a product that is
experimentally collected.
• The theoretical yield is the amount of a product that
should have formed.
– Theoretical Yield: Found from stoichiometry of the
LIMITING REAGENT
• Percent yield is a comparison of the actual amount of
product collected to what should have formed.
ActualYiel d
%Yield 
100%
TheoreticalYield
Sample Problem 12.10 (page 406)
Calcium carbonate, which is found in
seashells, is decomposed by heating. The
balanced equation for this reaction is:
CaCO3 (s)  CaO (s) + CO2 (g)
What is the theoretical yield of CaO if 24.8 g
CaCO3 is heated?
Sample Problem 12.11 (page 408)
What is the percent yield if 13.1 g CaO is
actually produced when 24.8 g CaCO3 is
heated?
CaCO3 (s)  CaO (s) + CO2 (g)
• More Difficult Problems: Mixing Limiting Reagent
with Percent Yield
What is the percent yield if 17.5 g of SO3 is
collected when 20.0 g of FeS2 is mixed with 16.0 g
O 2?
4FeS2 + 15O2  2Fe2O3 + 8SO3
• Identify your given information:
 17.5 g SO3 is the ____________________
 20.0 g FeS2 is _______________________
 16.0 g O2 is _______________________
• Find the LIMITING REAGENT:
• Find the THEORETICAL YIELD:
• Find the PERCENT YIELD:
• Find the LIMITING REAGENT:
 Choose one reactant to start with:
1molFeS2 15molO2 32.0 gO2
20.0 gFeS2
 20.0 gO2
120.0 gFeS2 4molFeS2 1molO2
 Since 20.0 g O2 is greater than the starting 16.0 g,
O2 is the LIMITING REAGENT
• Find the THEORETICAL YIELD
 SO3 is the Product we are trying to find
 Use 16.0 g O2 Limiting Reagent
1molO2 8molSO3 80.1gSO3
16.0 gO2
 21.4 gSO3
32.0 gO2 15molO2 1molSO3
 21.4 g SO3 is the Theoretical Yield
• Find the PERCENT YIELD
 Use the Theoretical Yield and the Actual Yield
17.5 gSO3
%Yield 
x100%  81.8%
21.4 gSO3