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Chemistry Chapter 9Chemistry Chapter 9

Unit 6Unit 6

StoichiometryStoichiometry

The arithmetic of equationsThe arithmetic of equations

�� Equations are recipes. Equations are recipes.

�� They tell chemists what amounts of reactants They tell chemists what amounts of reactants

to mix and what amounts of products to to mix and what amounts of products to

expect.expect.

��What is Stoichiometry?What is Stoichiometry?

�� The calculation of quantities in chemical The calculation of quantities in chemical

reactionsreactions

A quick exampleA quick example�� If we consider chemical equations to be recipes:If we consider chemical equations to be recipes:

�� When you bake cookies, you probably follow a recipeWhen you bake cookies, you probably follow a recipe

�� The recipe tells you what ingredients to mix and in The recipe tells you what ingredients to mix and in what ratiowhat ratio

�� The end result of the combination of ingredients is The end result of the combination of ingredients is cookiescookies

�� If you want more cookies you can double or triple the If you want more cookies you can double or triple the reciperecipe

�� If you want fewer cookies, the recipe can be cut in If you want fewer cookies, the recipe can be cut in half or quarteredhalf or quartered

�� The ingredients are the reactants and the cookies are The ingredients are the reactants and the cookies are the productsthe products

�� In a way, the cookie recipe gives the same kind In a way, the cookie recipe gives the same kind of information that a balanced chemical equation of information that a balanced chemical equation doesdoes

Stoichiometry: A Closer LookStoichiometry: A Closer Look

�� Lets examine Lets examine stoichiometrystoichiometry a little closer a little closer

by looking at the production of ammonia by looking at the production of ammonia

from it’s elements: Write the equation from it’s elements: Write the equation

below:below:

NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)

NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)

�� Valuable information can be derived from Valuable information can be derived from

this equation:this equation:

�� Write Write sentencessentences to describe the to describe the

equation equation --

1.1. In terms of PARTICLESIn terms of PARTICLES

One molecule of nitrogen gas reacts with One molecule of nitrogen gas reacts with

three molecules of hydrogen gas to three molecules of hydrogen gas to

produce two molecules of ammonia gas.produce two molecules of ammonia gas.

NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)

2.2. In terms of MOLESIn terms of MOLES

One mole of nitrogen gas reacts with One mole of nitrogen gas reacts with

three moles of hydrogen gas to produce three moles of hydrogen gas to produce

two moles of ammonia gas. two moles of ammonia gas.

The coefficients of a balanced chemical equation indicate The coefficients of a balanced chemical equation indicate

the relative number of moles of reactants to the moles the relative number of moles of reactants to the moles

of products in a reaction. This is the most important of products in a reaction. This is the most important

information obtained from a balanced chemical information obtained from a balanced chemical

equation.equation.

NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)

3.3. In terms of MASS In terms of MASS

Reactants:Reactants:

N = 14 x 2 = 28gN = 14 x 2 = 28g

H = 1 x 6 = +H = 1 x 6 = + 6g6g

34g total34g total

Products:Products:

N = 14 x 2 = 28gN = 14 x 2 = 28g

H = 1 x 6 = +H = 1 x 6 = + 6g6g

34g total34g total

Balanced chemical equations must obey the law of conservation ofBalanced chemical equations must obey the law of conservation ofmass. Remember that mass is related to number of atoms. Even mass. Remember that mass is related to number of atoms. Even though the number of moles is different, the mass of reactants though the number of moles is different, the mass of reactants

and products is equal.and products is equal.

NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)

4.4. In terms of VOLUME (for gases only!)In terms of VOLUME (for gases only!)

Remember that one mole of any gas at Remember that one mole of any gas at

STP is 22.4 L. Therefore, the volume (as STP is 22.4 L. Therefore, the volume (as

well as number of moles, molecules, and well as number of moles, molecules, and

formula units) of gases may not be the formula units) of gases may not be the

same.same.

NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)

�� What was conserved (stayed the same What was conserved (stayed the same

from left to right) in this equation? from left to right) in this equation?

(2 answers)(2 answers)

1. Mass1. Mass

2. Number of Atoms2. Number of Atoms

Another quick exampleAnother quick exampleImagine that you are in charge of manufacturing Imagine that you are in charge of manufacturing for Tiny Tyke Tricycle Company. The business for Tiny Tyke Tricycle Company. The business plan for Tiny Tike requires the production of 128 plan for Tiny Tike requires the production of 128 custom custom –– made tricycles each day. One of your made tricycles each day. One of your responsibilities is to be sure that there are responsibilities is to be sure that there are enough parts available at the start of each day to enough parts available at the start of each day to make these tricycles. To simplify this discussion, make these tricycles. To simplify this discussion, assume that the major components of the assume that the major components of the tricycle are the frame (F), the seat (S), the tricycle are the frame (F), the seat (S), the wheels (W), the handlebars (H), and the pedals wheels (W), the handlebars (H), and the pedals (P). The finished tricycle has a “formula” of (P). The finished tricycle has a “formula” of FSWFSW33HPHP22. The balanced equation for the . The balanced equation for the production of a tricycle is:production of a tricycle is:

F + S + 3W + H + 2P F + S + 3W + H + 2P →→ FSWFSW33HPHP22


�� In a five day workweek, Tiny Tike is scheduled In a five day workweek, Tiny Tike is scheduled

to make 640 tricycles. How many wheels should to make 640 tricycles. How many wheels should

be in the plant on Monday morning to make be in the plant on Monday morning to make

these tricycles?these tricycles?

�� # of tricycles to be made: 640 tricycles# of tricycles to be made: 640 tricycles

�� 1 FSW1 FSW33HPHP2 2 = 3W (from balanced equation)= 3W (from balanced equation)

�� Use the conversion factor: Use the conversion factor: 3W3W

1 FSW1 FSW33HPHP2 2

640 FSW640 FSW33HPHP2 2 x x 3W3W = 1920W= 1920W

1 FSW1 FSW33HPHP2 2


�� Here is another one:Here is another one:

�� How many tricycle seats, wheels, and pedals How many tricycle seats, wheels, and pedals

are needed to make 288 tricycles?are needed to make 288 tricycles?

Seats:Seats:

288288

Wheels:Wheels:

864 864

Pedals:Pedals:

576576

Practice Problems: Practice Problems: 1.1. Try this:Try this:

a.a. Write the equation for the complete Write the equation for the complete combustion of liquid ethanol, Ccombustion of liquid ethanol, C22HH55OH:OH:

b.b. Interpret the equation in terms of Interpret the equation in terms of numbers of moleculesnumbers of molecules

c.c. Interpret the equation in terms of Interpret the equation in terms of molesmoles

d.d. Interpret the equation in terms of massInterpret the equation in terms of mass

e.e. Why can’t this problem be interpreted Why can’t this problem be interpreted in terms of volume?in terms of volume?

Practice Problem 1Practice Problem 1

��Write the equation for the complete Write the equation for the complete

combustion of liquid ethanol, combustion of liquid ethanol,

CC22HH55OH:OH:

CC22HH55OHOH(l)(l) + 3O+ 3O2(g)2(g) →→ 3H3H22OO(l)(l) + 2CO+ 2CO2(g)2(g)


�� Interpret the equation in terms of Interpret the equation in terms of

numbers of molecules.numbers of molecules.

�� 1 molecule of liquid ethanol reacts with 1 molecule of liquid ethanol reacts with

3 molecules of oxygen gas to yield 3 3 molecules of oxygen gas to yield 3

molecules of liquid water and 2 molecules of liquid water and 2

molecules of carbon dioxide gas.molecules of carbon dioxide gas.


�� Interpret the equation in terms of Interpret the equation in terms of

molesmoles

�� 1 mole of liquid ethanol reacts with 3 1 mole of liquid ethanol reacts with 3

moles of oxygen gas to yield 3 moles of moles of oxygen gas to yield 3 moles of

liquid water and 2 moles of carbon liquid water and 2 moles of carbon

dioxide gas.dioxide gas.

�� 4 moles total reactants4 moles total reactants

�� 5 moles total products5 moles total products


�� Interpret the equation in terms of Interpret the equation in terms of massmass

�� Reactants:Reactants:

•• 1 molecule of ethanol = 46g1 molecule of ethanol = 46g

•• 3 molecules of oxygen = 96g3 molecules of oxygen = 96g

•• 142g total reactants142g total reactants

�� Products:Products:

•• 3 molecules of water 3 molecules of water = 54g= 54g

•• 2 molecules of carbon dioxide = 88g2 molecules of carbon dioxide = 88g

•• 142g total products142g total products


��Why can’t this problem be interpreted in Why can’t this problem be interpreted in

terms of volume? terms of volume?

�� Ethanol is in liquid form. The volume terms of Ethanol is in liquid form. The volume terms of

writing is based on gases.writing is based on gases.


��Write the balanced equation for the single Write the balanced equation for the single

–– replacement reaction of solid potassium replacement reaction of solid potassium

metal reacting with liquid water:metal reacting with liquid water:

2K2K(s)(s) + H+ H22OO(l)(l) →→ HH2(g)2(g) + K+ K22OO(aq)(aq)


�� Interpret this equation in terms of:Interpret this equation in terms of:

•• Interaction particlesInteraction particles--

•• molesmoles

•• massmass

•• volume (if allowed)volume (if allowed)


�� Interaction particles:Interaction particles:

�� Two atoms of solid potassium react with one Two atoms of solid potassium react with one

molecule of liquid water to produce one molecule of liquid water to produce one

molecule of hydrogen gas and one formula molecule of hydrogen gas and one formula

unit of aqueous potassium oxide.unit of aqueous potassium oxide.


��Moles:Moles:

�� Two moles of solid potassium metal reacts Two moles of solid potassium metal reacts

with one mole of liquid water to produce one with one mole of liquid water to produce one

mole of hydrogen gas and one mole of mole of hydrogen gas and one mole of

aqueous potassium oxide.aqueous potassium oxide.

�� 3 moles total reactants3 moles total reactants

�� 2 moles total products2 moles total products


��Mass:Mass:


•• K = 39g (2) = 78gK = 39g (2) = 78g

•• HH22O = 18g (1) = +O = 18g (1) = + 18g18g

96g total96g total


•• HH22 = 1g (2) = 2g= 1g (2) = 2g

•• KK22O = 94g (1) = +O = 94g (1) = + 94g94g

96g total96g total


�� volume (if allowed)volume (if allowed)

�� This is not allowed because not all are gases.This is not allowed because not all are gases.

Practice Test A Objective 1Practice Test A Objective 1

1.1. Write the balanced equation for the Write the balanced equation for the

formation of gaseous fluorine trioxide formation of gaseous fluorine trioxide

from its elements. Include the from its elements. Include the

“adjectives”.“adjectives”.

FF2(g)2(g) + 3O+ 3O2(g)2(g) →→ 2FO2FO3(g)3(g)


2.2. Interpret this equation in terms of the Interpret this equation in terms of the

number of representative particles.number of representative particles.

FF2(g)2(g) + 3O+ 3O2(g)2(g) →→ 2FO2FO3(g)3(g)

�� 1 molecule of fluorine gas reacts with 3 1 molecule of fluorine gas reacts with 3

molecules of oxygen gas to produce 2 molecules of oxygen gas to produce 2

molecules of fluorine trioxide gas.molecules of fluorine trioxide gas.


3.3. Interpret this equation in terms of number Interpret this equation in terms of number

of moles.of moles.

FF2(g)2(g) + 3O+ 3O2(g)2(g) →→ 2FO2FO3(g)3(g)

�� 1 mole of fluorine gas reacts with 3 moles of 1 mole of fluorine gas reacts with 3 moles of

oxygen gas to produce 2 moles of fluorine oxygen gas to produce 2 moles of fluorine

trioxide gas.trioxide gas.


4.4. Interpret this equation in terms of mass.Interpret this equation in terms of mass.

FF2(g)2(g) + 3O+ 3O2(g)2(g) →→ 2FO2FO3(g)3(g)


•• F = 19g (2) = 38gF = 19g (2) = 38g

•• O = 16g (6) = +O = 16g (6) = + 96g96g

134g total134g total


•• F = 19g (2) = 38gF = 19g (2) = 38g

•• O = 16g (6) = +O = 16g (6) = + 96g96g

134g total134g total


5.5. Interpret this equation in terms of Interpret this equation in terms of volume.volume.

FF2(g)2(g) + 3O+ 3O2(g)2(g) →→ 2FO2FO3(g)3(g)


•• FF22 = 1 mol x 22.4 L/mol = 22.4 L= 1 mol x 22.4 L/mol = 22.4 L

•• OO22 = 3 mol x 22.4 L/mol = + = 3 mol x 22.4 L/mol = + 67.2 L67.2 L

89.6 L total89.6 L total


•• FOFO33 = 2 mol x 22.4 L/mol = 44.8 L total= 2 mol x 22.4 L/mol = 44.8 L total

Chemical CalculationsChemical Calculations

Section 9.2Section 9.2

MoleMole--Mole Mole

CalculationsCalculations

The heart of the The heart of the stoichiometrystoichiometry

problem.problem.

Think WANT OVER GIVENThink WANT OVER GIVEN

�� Example Problem:Example Problem:

�� Rewrite the balanced equation for the Rewrite the balanced equation for the

production of ammonia from its elements:production of ammonia from its elements:

NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)

NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)

�� How many moles of ammonia are How many moles of ammonia are

produced when 0.6 mol of nitrogen react produced when 0.6 mol of nitrogen react

with hydrogen? with hydrogen?

0.6 mol N0.6 mol N22 x x 2 mol NH2 mol NH33 = 1.2 mol NH= 1.2 mol NH33

1mol N1mol N22

Example ProblemExample Problem

��Write the equation of aluminum oxide from Write the equation of aluminum oxide from

its elements.its elements.

4Al4Al(s)(s) + 3O+ 3O2(g)2(g) →→ 2Al2Al22OO3(s)3(s)


4Al4Al(s)(s) + 3O+ 3O2(g)2(g) →→ 2Al2Al22OO3(s)3(s)

1.1. How many moles of aluminum are needed to form 3.7 How many moles of aluminum are needed to form 3.7 moles of aluminum oxide?moles of aluminum oxide?

�� 3.7 mol 3.7 mol AlAl22OO3 3 x x 4mol Al4mol Al = 7.4 mol Al= 7.4 mol Al

2mol 2mol AlAl22OO33

2.2. How many moles of oxygen are required to react How many moles of oxygen are required to react completely with 14.8 moles of aluminum?completely with 14.8 moles of aluminum?

�� 14.8 mol Al x 14.8 mol Al x 3 mol O3 mol O22 = 11.1 mol O= 11.1 mol O22

4 mol Al4 mol Al

3.3. How many moles of aluminum oxide are formed when How many moles of aluminum oxide are formed when 0.78 moles of oxygen reacts with aluminum?0.78 moles of oxygen reacts with aluminum?

�� 0.78 mol O0.78 mol O22 x x 2 mol 2 mol AlAl22OO33 = 0.52 mol Al= 0.52 mol Al22OO33

3 mol O3 mol O22

MassMass--Mass CalculationsMass Calculations

�� Rewrite the equation for the production of Rewrite the equation for the production of

ammonia: ammonia:

NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)

�� Calculate the number of grams of ammonia Calculate the number of grams of ammonia

produced by the reaction of 5.4 grams of hydrogen produced by the reaction of 5.4 grams of hydrogen

with an excess of nitrogen.with an excess of nitrogen.

�� 5.4g H5.4g H22 x x 1 mol H1 mol H22 = 2.7 mol H= 2.7 mol H22

2g H2g H22

�� 2.7 mol H2.7 mol H22 x x 2 mol NH2 mol NH3 3 = 1.8 mol NH= 1.8 mol NH33

3 mol H3 mol H22

�� 1.8 mol NH1.8 mol NH3 3 x x 17.0g NH17.0g NH33 = 31g NH= 31g NH33

1 mol NH1 mol NH33

Example Problem Example Problem

�� Write the balanced chemical equation for the Write the balanced chemical equation for the decomposition for potassium chlorate.decomposition for potassium chlorate.

�� Calculate the number of grams of reactant Calculate the number of grams of reactant necessary to produce 50 grams of potassium necessary to produce 50 grams of potassium chloride.chloride.

�� How many grams of oxygen gas are produced How many grams of oxygen gas are produced when you decompose 15 grams of potassium when you decompose 15 grams of potassium chlorate?chlorate?

�� How many MOLES of each product are How many MOLES of each product are produced when 10 grams of potassium chlorate produced when 10 grams of potassium chlorate are heated?are heated?

�� How many grams of each product are produced How many grams of each product are produced when 2.5 moles of reactant are used? when 2.5 moles of reactant are used?


��Write the balanced chemical equation for Write the balanced chemical equation for

the decomposition for potassium chlorate.the decomposition for potassium chlorate.

2KClO2KClO33 →→ 3O3O22 + 2KCl+ 2KCl


�� Calculate the number of grams of reactant Calculate the number of grams of reactant

necessary to produce 50 grams of necessary to produce 50 grams of

potassium chloride.potassium chloride.

50g 50g KClKCl x x 1 mol 1 mol KClKCl x x 2 mol KClO2 mol KClO33 x x 122.5g KClO122.5g KClO3 3 = 82.21g KClO= 82.21g KClO33

74.5g 74.5g KClKCl 2 mol 2 mol KClKCl 1 mol KClO1 mol KClO33


�� How many grams of oxygen gas are How many grams of oxygen gas are

produced when you decompose 15 grams produced when you decompose 15 grams

of potassium chlorate?of potassium chlorate?

15g KClO15g KClO3 3 x x 1mol KClO1mol KClO3 3 x x 3mol O3mol O22 x x 32g O32g O2 2 = 5.88g O= 5.88g O22

122.5g KClO122.5g KClO3 3 2mol KClO2mol KClO33 1mol O1mol O22


�� How many MOLES of each product are How many MOLES of each product are

produced when 10 grams of potassium produced when 10 grams of potassium

chlorate are heated?chlorate are heated?

10g KClO10g KClO3 3 x x 1mol KClO1mol KClO3 3 x x 3mol O3mol O22 = 0.12mol O= 0.12mol O22

122.5g KClO122.5g KClO3 3 2mol KClO2mol KClO33

10g KClO10g KClO3 3 x x 1mol KClO1mol KClO3 3 x x 2mol 2mol KClKCl = 0.08mol = 0.08mol KClKCl

122.5g KClO122.5g KClO3 3 2mol KClO2mol KClO3 3


�� How many grams of each product are How many grams of each product are produced when 2.5 moles of reactant are produced when 2.5 moles of reactant are used? used?

2.5 mol KClO2.5 mol KClO3 3 x x 3 mol O3 mol O22 = 3.75 mol O= 3.75 mol O22

2 mol KClO2 mol KClO33

3.75 mol O3.75 mol O22 x x 32g O32g O22 = = 120g O120g O22

1 mol O1 mol O22

2.5 mol KClO2.5 mol KClO3 3 x x 2 mol 2 mol KClKCl = 2.5 mol = 2.5 mol KClKCl

2 mol KClO2 mol KClO33

2.5 mol 2.5 mol KClKCl x x 74.5g 74.5g KClKCl = = 186.25g 186.25g KClKCl

1 mol 1 mol KClKCl

Other Other StoichiometricStoichiometric

CalculationsCalculations

The The StoichiometryStoichiometry Road MapRoad Map

��With your knowledge of conversion factors With your knowledge of conversion factors

and the problemand the problem--solving diagram, you can solving diagram, you can

solve a variety of solve a variety of stoichiometricstoichiometric problems.problems.

And Have lots of fun doing it! And Have lots of fun doing it!


�� How many molecules of oxygen are How many molecules of oxygen are

produced when a sample of 29.2 grams of produced when a sample of 29.2 grams of

water is decomposed by electrolysis?water is decomposed by electrolysis?

Balanced Equation: 2HBalanced Equation: 2H22OO(l)(l) →→ 2H2H22(g)(g) + O+ O2(g)2(g)

29.2g H29.2g H22O x O x 1mol H1mol H22OO X X 1mol O1mol O22 X X 6.02 e^6.02 e^2323 molecules Omolecules O22 ==

18g H18g H220 2mol H0 2mol H22O 1 mol 0O 1 mol 022

4.8 X 10^4.8 X 10^2323 molecules Omolecules O22


�� How many liters of oxygen are produced How many liters of oxygen are produced

by the decomposition of 6.54 grams of by the decomposition of 6.54 grams of

potassium chlorate?potassium chlorate?

Balanced Equation: 2KClOBalanced Equation: 2KClO3(s)3(s) →→ 3O3O2(g)2(g) + 2KCl+ 2KCl(s)(s)

6.54g KClO6.54g KClO33 x x 1mol KClO1mol KClO33 x x 3mol O3mol O22 x x 22.4L O22.4L O22 ==

122.5g KClO122.5g KClO33 2mol KClO2mol KClO3 3 1mol O1mol O22

1.79L O1.79L O22


�� Assuming STP, how many liters of oxygen gas Assuming STP, how many liters of oxygen gas

are needed to produce 19.8 L sulfur trioxide are needed to produce 19.8 L sulfur trioxide

according to this balanced equation?according to this balanced equation?

2SO2SO2(g)2(g) + O+ O2(g)2(g) �� 2SO2SO3(g)3(g)

19.8L SO19.8L SO33 x x 1mol SO1mol SO33 x x 1mol O1mol O22 x x 22.4L O22.4L O22 ==

22.4L SO22.4L SO3 3 2mol SO2mol SO3 3 1mol O1mol O22

9.9L O9.9L O22


�� Nitrogen monoxide and oxygen gas combine to form the Nitrogen monoxide and oxygen gas combine to form the

brown gas nitrogen dioxide. How many milliliters of brown gas nitrogen dioxide. How many milliliters of

nitrogen dioxide are produced when 3.4 milliliters of nitrogen dioxide are produced when 3.4 milliliters of

oxygen react with an excess of nitrogen monoxide? oxygen react with an excess of nitrogen monoxide?

Assume STP conditions. Assume STP conditions.

Balanced Equation: 2NO + OBalanced Equation: 2NO + O22 →→ 2NO2NO22

0.0034L O0.0034L O22 x x 1mol O1mol O22 x x 2mol NO2mol NO22 x x 22.4L NO22.4L NO22 x x 1000ml NO1000ml NO2 2 ==

22.4L O22.4L O22 1mol O1mol O22 1mol NO1mol NO22 1L NO1L NO22

Practice ProblemsPractice Problems

See Note Pack and TransparenciesSee Note Pack and Transparencies

And most of all…..And most of all…..

Have FunHave Fun

Section 9.3Section 9.3

Limiting ReagentLimiting Reagent

Limiting ReagentLimiting Reagent

�� The limiting reagent limits or determines The limiting reagent limits or determines

the amount of product that can be formed the amount of product that can be formed

in a reactionin a reaction

�� In contrast, the excess reagent is the In contrast, the excess reagent is the

reagent that is not used up in a reaction reagent that is not used up in a reaction

(there is some left over)(there is some left over)

Ammonia Production (again)Ammonia Production (again)

NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)

�� This represents the most efficient recipe This represents the most efficient recipe

that a chemist can followthat a chemist can follow

��What is the limiting reagent if the reaction What is the limiting reagent if the reaction

is run with 2 moles of nitrogen and 3 is run with 2 moles of nitrogen and 3

moles of hydrogen?moles of hydrogen?

HYDROGENHYDROGEN


�� Sodium chloride can be prepared by the reaction Sodium chloride can be prepared by the reaction

of sodium with chlorine gas according to the of sodium with chlorine gas according to the

following equation:following equation:

2Na2Na(s)(s) + Cl+ Cl2(g)2(g) →→ 2NaCl2NaCl

�� Suppose that 6.7 mol of sodium reacts with 3.2 Suppose that 6.7 mol of sodium reacts with 3.2

mol of chlorine.mol of chlorine.

�� How many moles of sodium chloride are produced?How many moles of sodium chloride are produced?

�� What is the limiting reagent?What is the limiting reagent?

Follow along on page 254 of the bookFollow along on page 254 of the book


2Na2Na(s)(s) + Cl+ Cl2(g)2(g) →→ 2NaCl2NaCl

�� Suppose that 6.7 mol of sodium reacts with 3.2 Suppose that 6.7 mol of sodium reacts with 3.2

mol of chlorine.mol of chlorine.

�� How many moles of sodium chloride are produced?How many moles of sodium chloride are produced?

Choose one known and compare it to the required amount of the otChoose one known and compare it to the required amount of the other reactant.her reactant.

6.7 mol Na x 1mol Cl6.7 mol Na x 1mol Cl22 / 2 mol Na = 3.35 mol Cl/ 2 mol Na = 3.35 mol Cl2 2 (Limiting)(Limiting)

3.20 mol Cl3.20 mol Cl22 x 2 mol x 2 mol NaClNaCl / 1 mol Cl/ 1 mol Cl22 = 6.40 mol = 6.40 mol NaClNaCl (Answer)(Answer)

�� What is the limiting reagent?What is the limiting reagent?

The limiting reagent becomes ClThe limiting reagent becomes Cl22 because we do not because we do not

have enough to completely react all of the Nahave enough to completely react all of the Na

Example Problem 2Example Problem 2

��Write the equation for the complete Write the equation for the complete

combustion of combustion of etheneethene, C, C22HH44

�� If 2.7 mol of If 2.7 mol of etheneethene is reacted with 6.3 mol is reacted with 6.3 mol

of oxygen…of oxygen…

�� Calculate the moles of water producedCalculate the moles of water produced

�� What is the limiting reagentWhat is the limiting reagent


��Write the equation for the incomplete Write the equation for the incomplete

combustion of combustion of etheneethene, C, C22HH44

�� If 2.7 mol of If 2.7 mol of etheneethene is reacted with 6.3 mol is reacted with 6.3 mol

of oxygen…of oxygen…

�� Calculate the moles of water producedCalculate the moles of water produced

�� What is the limiting reagentWhat is the limiting reagent


��When copper reacts with sulfur, solid When copper reacts with sulfur, solid

copper (I) sulfide is produced. Write this copper (I) sulfide is produced. Write this

reaction.reaction.

��What is the limiting reagent when 80.0 What is the limiting reagent when 80.0

grams of copper reacts with 25.0 grams of grams of copper reacts with 25.0 grams of

sulfur?sulfur?

��What is the maximum number of grams of What is the maximum number of grams of

product that can be formed?product that can be formed?


�� Hydrogen gas can be produced in the lab by the Hydrogen gas can be produced in the lab by the

reaction of magnesium metal with hydrochloric reaction of magnesium metal with hydrochloric

acid. Write the COMPLETE balanced equation.acid. Write the COMPLETE balanced equation.

�� Identify the limiting reagent when 6 grams of Identify the limiting reagent when 6 grams of HClHCl

react with 5 grams of Mg.react with 5 grams of Mg.

�� How many grams of hydrogen can be How many grams of hydrogen can be

produced?produced?

The EndThe End

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Chemistry Chapter 9 note ppt - Reeths-Puffermoodle.reeths-puffer.org/pluginfile.php/9601/mod_page/content/2... · Chemistry Chapter 9 Unit 6 Stoichiometry. ... How many moles of aluminum