The Study of Stoichiometry I. Stoichiometric Calculations.

The Study of The Study of StoichiometryStoichiometryThe Study of The Study of

StoichiometryStoichiometry

I. I. Stoichiometric Stoichiometric CalculationsCalculations

A. Proportional A. Proportional RelationshipsRelationshipsA. Proportional A. Proportional RelationshipsRelationships

How many eggs are needed to make 12 dozen cookies?

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

12 doz. 2 eggs

5 doz.= 5 eggs

Ratio of eggs to cookies


StoichiometryStoichiometry• calculating amounts of reactants &

products using mole ratios

Mole RatioMole Ratio• indicated by coefficients in a

balanced equation

2 Mg + O2 Mg + O22 2 MgO 2 MgO


Mole Ratio examples:Mole Ratio examples:

2 H2 H22 + O + O22 2 H 2 H22OO2 mol H2 mol H22 : 1 mol O : 1 mol O22 2 mol H 2 mol H22OO

CHCH4 4 + 2O+ 2O22 CO CO22 + 2H + 2H22OO1 mol CH1 mol CH4 4 : 2 mol O: 2 mol O22 1 mol CO1 mol CO2 2 : 2 mol H: 2 mol H22OO

B. Stoichiometry StepsB. Stoichiometry StepsB. Stoichiometry StepsB. Stoichiometry Steps

1. Write a balanced equation.

2. Identify known & unknown.

3. Line up conversion factors.• Mole ratio - moles moles• Molar mass - moles grams

Core step in all stoichiometry problems!!

• Mole ratio - moles moles

4. Check answer.

C. Stoichiometry C. Stoichiometry ProblemsProblemsC. Stoichiometry C. Stoichiometry ProblemsProblems

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


How many grams of KCl will be formed from 2.5 mol KClO3?

2.5 mol KClO3

2 molKCl

2 molKClO3

= 186.3 g KCl

2KClO3 2KCl + 3O2

74.55 gKCl

1 molKCl

2.5 mol ? g


How many grams of silver will be formed from 12.0 g copper?

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


1. Try on your ownHow many grams of calcium carbonate are required to prepare 50.0 g of calcium oxide?

CaCO3 CaO + CO2

II. Gas StoichiometryII. Gas Stoichiometry

Ch. 5 - StoichiometryCh. 5 - StoichiometryCh. 5 - StoichiometryCh. 5 - Stoichiometry

1 mol of a gas=22.4 Lat STP

A. Molar Volume at STPA. Molar Volume at STPA. Molar Volume at STPA. Molar Volume at STP

Standard Temperature & Pressure0°C and 1 atm

A. Molar Volume at STPA. Molar Volume at STPA. Molar Volume at STPA. Molar 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)

B. Gas Stoichiometry B. Gas Stoichiometry ProblemProblemB. Gas Stoichiometry B. Gas Stoichiometry ProblemProblem

How many grams of CaCO3 are req’d to

produce 9.00 L of CO2 at STP?

9.00 LCO2

1 molCO2

22.4 L CO2

= 40.2 g CaCO3

CaCO3 CaO + CO2

1 molCaCO3

1 molCO2

100.09g CaCO3

1 molCaCO3

? g 9.00 L

StoichiometryStoichiometryStoichiometryStoichiometry

Limiting Reactants

A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting 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 ReactantLimiting Reactant• used up in a reaction• determines the amount of product

Excess ReactantExcess Reactant• added to ensure that the other

reactant is completely used up• cheaper & easier to recycle


1. Write a balanced equation.

2. For each reactant, calculate the

amount of product formed.

3. Smaller answer indicates:

• limiting reactant

• amount of product


79.1 g of zinc react with 2.1 mol HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP?

Zn + 2HCl ZnCl2 + H2 79.1 g ? L2.1 mol


79.1g Zn

1 molZn

65.39g Zn

= 27.1 L H2

1 molH2

1 molZn

22.4 LH2

1 molH2



22.4L H2

1 molH2

2.1 mol HCl

1 mol H2

2 mol HCl

= 23.52 L H2



Zn: 27.1 L H2 HCl: 23.5 L H2

Limiting reactant: HCl

Excess reactant: Zn

Product Formed: 23.52 L H2

left over zinc


35.6 g of zinc react with 1.9 mol HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP?



35.6g Zn

1 molZn

65.39g Zn

= 12.2 L H2

1 molH2

1 molZn

22.4 LH2

1 molH2



22.4L H2

1 molH2

1.9 mol HCl

1 mol H2

2 mol HCl

= 21.28 L H2



Zn: 12.2 L H2 HCl: 21.28 L H2

Limiting reactant: Zn

Excess reactant: HCl

Product Formed: 12.2 L H2

left over HCl

B. Percent YieldB. Percent YieldB. Percent YieldB. Percent Yield

100yield ltheoretica

yield actualyield %

calculated on paper

measured in lab


When 45.8 g of K2CO3 react with excess

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

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

actual: 46.3 g


45.8 gK2CO3

1 molK2CO3

138.21 gK2CO3

= 49.4g KCl

2 molKCl

1 molK2CO3

74.55g KCl

1 molKCl

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

actual: 46.3 g

Theoretical Yield:


Theoretical Yield = 49.4 g KCl

% Yield =46.3 g

49.4 g 100 =93.7%

K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g 49.4 g

actual: 46.3 g

2 HCl + Zn H2 + ZnCl2130.7 g Zn was able to produce

twice its mass of Zinc Chloride in the chemical reaction above. What

is the percent yield of ZnCl2?

=130.7 1

65.39

g Zn Mol Zn

g Zn272.4 g ZnCl2

g ZnCl2136.28

mol ZnCl2

1

Percent Yield

1 mol Zn

mol ZnCl21

= Actual

Calculated

261.4 g ZnCl2

272.4 g ZnCl2

X 100= 95.9% Yield

HCl + NaOH NaCl + H NaCl + H2O

100 g HCl reacts with 100 g NaOH in the chemical reaction above.

What is the limiting reactant and how many grams of NaCl will be

produced by the reaction?

=100 1

36.45

g HCl mol HCl

g HCl

160.3 g NaCl

g NaCl58.44

mol NaCl1

=100 1

40.0

g NaOH

mol NaOHg NaOH

146.1 g NaCl

mol NaCl1

mol NaOH

1 mol NaCl

1

g NaCl58.44

Limiting Reactant

1 mol HCl

mol NaCl1

The Study of Stoichiometry I. Stoichiometric Calculations.

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