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I. Stoichiometric
Calculations
I. Stoichiometric
Calculations
Stoichiometry – Ch. Stoichiometry – Ch. 1111
Background on things you Background on things you NEED to know how to do:NEED to know how to do:Background on things you Background on things you NEED to know how to do:NEED to know how to do:
1. Name/write correct chemical formula
2. Write chemical equations
3. Balance chemical equations
4. Predict Products
5. Mole/mass conversions
StoichiometryStoichiometryStoichiometryStoichiometryo Stoichiometry uses ratios to determine relative
amounts of reactants or products.o For example If you were to make a bicycle,
you would need one frame and two tires.o 1 frame + 2 tires 1 bicycleo If I had 74 tires, what is the most # of bicycles
I could make?
74 tires
2 tires
1 bicycle = 37 bicycles
Proportional Proportional RelationshipsRelationshipsProportional Proportional RelationshipsRelationships
I have 5 eggs. How many cookies can I make?
3/4 c. brown sugar1 tsp vanilla extract2 eggs2 c. chocolate chipsMakes 5 dozen cookies.
2 1/4 c. flour1 tsp. baking soda1 tsp. salt1 c. butter3/4 c. sugar
5 eggs 5 doz.
2 eggs= 12.5 dozen cookies
Ratio of eggs to cookies
Proportional Proportional RelationshipsRelationshipsProportional Proportional RelationshipsRelationships
StoichiometryStoichiometry• mass relationships between substances in a
chemical reaction• based on the mole ratio
Mole RatioMole Ratio• indicated by coefficients in a balanced equation• can be used to determine expected amounts of
products given amounts of reactants.
2 Mg + O2 Mg + O22 2 MgO 2 MgO
Stoichiometry StepsStoichiometry StepsStoichiometry StepsStoichiometry Steps
1. Write a balanced equation.2. Identify known & unknown.3. Line up conversion factors.
• Mole ratio - moles moles• Molar mass - moles grams• Molarity - moles liters soln• Molar volume - moles liters gas
Core step in all stoichiometry problems!!
• Mole ratio - moles moles
4. Check answer.
Molar Volume at STPMolar Volume at STPMolar Volume at STPMolar Volume at STP
Molar Mass(g/mol)
6.02 1023
particles/mol
MASSIN
GRAMSMOLES
NUMBEROF
PARTICLES
LITERSOF
SOLUTION
Molar Volume (22.4 L/mol)
LITERSOF GASAT STP
Molarity (mol/L)
Mole – Mole ConversionsMole – Mole ConversionsMole – Mole ConversionsMole – Mole Conversions The first type of problems we encounter will
go between moles and moles. For this we need to use mole ratios.
Ex: Write and balance the reaction between lead (II) nitrate and potasium iodide.
Pb(NO3)2 + 2KI 2 KNO3 + PbI2
Mole ratio of potasium iodide to lead (II) iodide:
2 moles KI
1 mole PbI2
Mole to Mole ProblemsMole to Mole ProblemsMole to Mole ProblemsMole to Mole Problems
How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?
9 mol O2 2 mol KClO3
3 mol O2
= 6 mol KClO3
2KClO3 2KCl + 3O2 ? mol 9 mol
Mole to MassMole to MassMole to MassMole to Mass
We can also convert from moles to mass, and mass to moles
For Example:
• 4 Al + 3 O2 2Al2O3
If you know how many grams of Al you start with, we can write a flow chart to show how to calculate the # of moles of oxygen need to fully react with the Al.
g Al moles Al moles of oxygen
Mass to Moles: Mass to Moles: 4 Al + 3 O4 Al + 3 O22 2Al2Al22OO33
Mass to Moles: Mass to Moles: 4 Al + 3 O4 Al + 3 O22 2Al2Al22OO33
If the reaction starts with .84 moles of aluminum, how many grams of aluminum oxide are produced?
.84 mol Al
4 mol Al
2 mol Al2O3
1 mol Al2O3
101.9 grams Al2O3
= 42.8 grams Al2O3 0.92 g of Aluminum oxide are produced from the
reaction. How much aluminum was used up?
.92 g Al2O3
101.9 g Al2O3
1 mol Al2O3
2 mol Al2O3
4 mol Al
1 mol Al
26.9 g Al
= .49 grams Al
Mass to MassMass to MassMass to MassMass to Mass How many grams of silver will be
formed from 12.0 g copper reacting with silver nitrate?
12.0g Cu
1 molCu
63.55g Cu
= 40.7 g Ag
Cu + 2AgNO3 2Ag + Cu(NO3)2
2 molAg
1 molCu
107.87g Ag
1 molAg
12.0 g ? g
Stoichiometry Problems – Mole/MassStoichiometry Problems – Mole/MassStoichiometry Problems – Mole/MassStoichiometry Problems – Mole/Mass
In photosynthesis, carbon dioxide and water react to form glucose, C6H12O6 and oxygen gas.
___CO2 + ___H2O ___C6H12O6 + ___O2
If 15.6 grams of carbon dioxide react, how many moles of glucose will be produced?
How many grams of carbon dioxide must react to produce 0.25 moles of glucose?
6 6 6
15.6 g CO2
44.01 g CO2
1 mol CO2
6 mol CO2
1 mol C2 H12O6 = 0.0591 mol C2H12O6
44.01 g CO26 mol CO2
1 mol CO2
0.25 mol C2 H12O6= 66 g
CO21 mol C2 H12O6
Stoichiometry with GasesStoichiometry with GasesStoichiometry with GasesStoichiometry with Gases
If the pressure and temperature are constant, the ratio of moles in the balanced equation is the ratio of liters in an all gas reaction.
At STP 1 mol of a gas=22.4 L
Molar Volume at STPMolar Volume at STPMolar Volume at STPMolar Volume at STP
Standard Temperature & Pressure0°C and 1 atm
Molar VolumeMolar VolumeMolar VolumeMolar Volume
Hydrogen and chlorine gas react to produce hydrochloric acid. If 7.00 L of hydrogen gas react, how many liters of HCl gas are formed?
H2 (g) + Cl2 (g) 2 HCl (g)
7.00 L H2
1.0 L H2
2.0 L HCl = 14.0 L HCl
*only in all gas reactions!
Molar VolumeMolar VolumeMolar VolumeMolar Volume
In the following reaction, if 17 g of Mg react, how many L of H2 forms?
Mg (s) + 2HCl (aq) MgCl2 (aq) + H2 (g)
17.0 gMg
1 molMg
24.31 gMg
1 molH2
1 molMg
22.4 L H2
1 molH2
= 15.7 L H2
How many grams of KClO3 are req’d to
produce 9.00 L of O2 at STP?
9.00 LO2
1 molO2
22.4 L O2
= 32.8 g KClO3
2 molKClO3
3 molO2
122.55g KClO3
1 molKClO3
? g 9.00 L
Molar Volume ProblemsMolar Volume ProblemsMolar Volume ProblemsMolar Volume Problems
2KClO3 2KCl + 3O2
63.55g Cu
1 molCu
Molar Volume ProblemsMolar Volume ProblemsMolar Volume ProblemsMolar Volume Problems
How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3?
1.5L
.10 molAgNO3
1 L= 4.8 g
Cu
Cu + 2AgNO3 2Ag + Cu(NO3)2
1 molCu
2 molAgNO3
? g 1.5L0.10M
MolarityMolarityMolarityMolarity
Molarity is the number of moles of solute dissolved in one liter of solution.
Units are moles per liter or moles of solute per liter of solution.
Molarity abbreviated by a capital M Molarity = moles of solute
liter of solution
Molarity ProblemsMolarity ProblemsMolarity ProblemsMolarity Problems
As an example, suppose we dissolve 23 g of ammonium chloride (NH4Cl) in enough water to make 145 mL of solution. What is the molarity of ammonium chloride in this solution?
23 g NH4Cl
53.5 g NH4Cl
1 mole NH4Cl = .43 mol NH4Cl
145 mL
1000 mL
1 L = .145 L
.43 mol NH4Cl
.145 L= 2.97 M NH4Cl
Molarity ProblemsMolarity ProblemsMolarity ProblemsMolarity Problems Now, suppose we have a beaker with 175 mL
of a 5.5 M HCl solution. How many moles of HCl is in this beaker?
1000 mL
1 L = .96 mol HCl175 mL
1 L
5.5 mol HCl
Suppose you had 70 grams of NaCl and you dissolved it in exactly 2.00 L of solution. What would be the molarity of the solution?
70 g NaCl
58.44 g NaCl
1 mol NaCl = 1.2 mol
NaCl
1.2 mol NaCl 2.0 L
= 0.6 M NaCl
Volume of SolutionsVolume of SolutionsVolume of SolutionsVolume of Solutions
A 10.% HCl solution (soln) means:
10 g HCl (pure) 100 g HCl soln
A solution with a density of 1.5 g/mL means: 1.5 g soln 1 mL soln
Impure SubstancesImpure SubstancesImpure SubstancesImpure Substances To say a substance is 75% NaCl by mass means:
75 g NaCl (pure) 100 g of NaCl solution
or
100 g of NaCl solution
75 g of NaCL (pure) Or, Iron ore that is 15% iron by mass means:
15 g Fe 100 g ore
or
100 g of ore100 g of ore
15 g of Fe15 g of Fe
Exothermic and EndothermicExothermic and EndothermicExothermic and EndothermicExothermic and Endothermic
Exothermic process – heat is released into the surroundings
• Exo = Exit
Endothermic Process – heat is absorbed from the surroundings
• Endo = Into
HEAT
HEAT
Thermochemical EquationsThermochemical EquationsThermochemical EquationsThermochemical Equations
In a thermochemical equation, the energy of change for the reaction can be written as either a reactant or a product
Enthalpy: the heat content of a system at constant pressure (ΔH)
Endothermic (positive ΔH)
2NaHCO3 + 129kJ Na2CO3 + H2O + CO2
Exothermic (negative ΔH)
CaO + H2O Ca(OH)2 + 65.2kJ
Write the thermochemical equation for the oxidation of Iron (III) if its ΔH= -1652 kJ
Fe(s) + O2(g)→ Fe2O3(s) + 1652 kJ
How much heat is evolved when 10.00g of Iron is reacted with excess oxygen?
4 3 2
10.00g Fe
55.85g Fe
1 mol Fe
4 mol Fe
1652 kJ=73.97 kJ of heat
Exo
Write the thermochemical equation for the decomposition of sodium bicarbonate, with a ΔH = + 129 kJ:
2 NaHCO3 + 129kJ → Na2CO3(s) + H2O + CO2
How much heat is required to break down 50.0g of sodium bicarbonate?
50.0 g NaHCO3
83.9 g NaHCO3
1 mol NaHCO3
2 mol NaHCO3
129 kJ
=38.4 kJ of heat
Endo
Write the thermochemical equation for the synthesis of calcium oxide and water with a ΔH= - 65.2 kJ:
CaO + H2O → Ca(OH)2 + 65.2kJ
How much energy is released when 100 g of calcium oxide reacts?
100 g CaO
56.07 g CaO
1 mol CaO
1 mol CaO
65.2 kJ
=116 kJ of heat
Exo
Write the thermochemical equation for the decomposition of magnesium oxide with a ΔH= + 61.5 kJ:
2 MgO + 61.5 → 2 Mg + O2
How many grams of oxygen are produced when magnesium oxide is decomposed by adding 420 kJ of Energy?
420 kJ
61.5 kJ
1 mol O2
1 mol O2
31.98 g O2 =218 g of O2
Endo
Stoichiometry in the Real World
Stoichiometry in the Real World
Stoichiometry – Ch. Stoichiometry – Ch. 1111
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• 24 graham cracker squares• 1 bag of marshmallows• 12 pieces of chocolate
Limiting ReactantLimiting Reactant• chocolate
Excess ReactantsExcess Reactants• Marshmallows and
graham crackers
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants
Limiting ReactantLimiting Reactant• one that is used up in a reaction• determines the amount of product that
can be produced
Excess ReactantExcess Reactant• added to ensure that the other reactant
is completely used up• cheaper & easier to recycle
Limiting Reactant StepsLimiting Reactant StepsLimiting Reactant StepsLimiting Reactant Steps
1. Write the balanced equation.
2. For each reactant, calculate the
amount of product formed.
3. Smaller answer indicates:
• limiting reactant
• amount of product actually possible
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants
79.1 g of zinc react with 68.1 g HCl. Identify the limiting and excess reactants. How many grams of hydrogen can be formed?
Zn + 2HCl ZnCl2 + H2 79.1 g ? g68.1 g
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants
2.02 g H2
1 molH2
68.1g HCl
1 molHCl
36.46 g HCl
= 1.89 g H2
1 molH2
2 molHCl
Zn + 2HCl ZnCl2 + H2 79.1 g ? g68.1 g
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants
79.1g Zn
1 molZn
65.39g Zn
= 2.44 g H2
1 molH2
1 molZn
2.02 gH2
1 molH2
Zn + 2HCl ZnCl2 + H2 79.1 g ? g68.1 g
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants
Zn: 2.44 g H2 HCl: 1.89 g H2
Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 1.89 g H2
left over zinc
Limiting Reactants #2Limiting Reactants #2 Limiting Reactants #2Limiting Reactants #2
5.42 g of magnesium ribbon react with 4.00 g of oxygen gas. Identify the limiting and excess reactants. How many grams of magnesium oxide are formed?
2Mg + O2 2MgO 5.42 g ? g4.00 g
Limiting Reactants #2Limiting Reactants #2Limiting Reactants #2Limiting Reactants #2
5.42g Mg
1 molMg
24.31g Mg
= 8.99 g MgO
2 molMgO
2 molMg
40.31 gMgO
1 molMgO
2Mg + O2 2MgO
5.42 g ? g4.00 g
Limiting Reactants #2Limiting Reactants #2Limiting Reactants #2Limiting Reactants #2
40.31g MgO
1 molMgO
4.00g O2
1 molO2
32.00 g O2
= 10.1 g MgO
2 molMgO
1 molO2
2Mg + O2 2MgO
5.42 g ? g4.00 g
A. Limiting Reactants #2A. Limiting Reactants #2A. Limiting Reactants #2A. Limiting Reactants #2
Mg: 8.99 g MgO O2: 10.1 g MgO
Excess oxygen
Limiting reactant: Mg
Excess reactant: O2
Product Formed: 8.99 g MgO
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants
What other information could you find in these problems?
• How much of each reactant is used – in grams, liters, moles
• How much of excess reactant is left over – in grams, liters, moles
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 and % yields of KCl.
1 molK2CO3
138.2 g K2CO3
2 molKCl
1 molK2CO3
74.55 gKCl
1 molKCl
45.8 gK2CO3
K2CO3 + 2 HCl 2 KCl + H2CO3
= 49.4 grams KCl