Chapter 4 Quantities of Reactants and Productsmtweb.mtsu.edu/nchong/Chapter 4-CHEM1010-MSJ.pdf ·...

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John W. MooreConrad L. StanitskiPeter C. Jurs

Stephen C. Foster • Mississippi State University

http://academic.cengage.com/chemistry/moore

Chapter 4Quantities of Reactants and Products

ReactantsReactants ProductsProducts

C6H12O6(aq) 2 C2H5OH(ℓ) + 2 CO2(g)glucose ethanol carbon dioxide

yeast

Conditions may be shown over the arrow. e.g.

heat (Δ) reflux catalyst present (yeast)

Physical states are often listed:

(g) gas (s) solid(ℓ) liquid (aq) aqueous (dissolved in water)

Chemical Equations

O2 molar mass

Balanced equationsBalanced equations obey the law of conservation ofmass (Lavoisier 1789).

2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(ℓ)

NanoscaleNanoscale 2 molecules 7 molecules 4 molecules 6 molecules

MacroscaleMacroscale 2 moles 7 moles 4 moles 6 moles

2(30.0)= 60.0 g 7(32.0)= 224.0 g 4(44.0)= 176.0 g 6(18.0)= 108.0 g

C2H6 molar mass

284.0 g 284.0 g

“Mass is neither created nor destroyed in a chemical reaction.”

Chemical EquationsStoichiometryStoichiometry

The relationship between the number of reactant andproduct molecules in a chemical equation.

CaCO3(s) + 2 HNO3(aq)

Ca(NO3)2(aq) + CO2(g) + H2O(ℓ)A stoichiometric

coefficient

Chemical Equations

Combination Reactions

+

X Z XZ

Element plus halogen or O2:

2 Mg(s) + O2(g) 2 MgO(s)

I2(s) + Zn(s) ZnI2(s)

There are other types:

2 SO2(g) + O2(g) 2 SO3(g)

+

XZ X Z

Occasionally by shock:

4 C3H5(NO3)3(ℓ)

12 CO2(g) + 10 H2O(ℓ) + 6 N2(g) + O2(g)

Often initiated by heat:

CaCO3(s) CaO(s) + CO2(g)

2 KNO3(s) 2 KNO2(s) + O2(g)heat

800 - 1000°C

Decomposition Reactions

nitroglycerin

2

Decomposition Reactions Displacement Reactions

+ +

A XZ AZ X

Metals may displace another metal from its salt

Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)

Zn(s) + 2 AgNO3(aq) → Zn(NO3)2(aq) + 2 Ag(s)

Some other examples:

F2(g) + 2 LiCl(s) → 2 LiF(s) + Cl2(g)

2 Na(s) + 2 H2O(ℓ) → 2 NaOH(aq) + H2(g)

Displacement Reactions Exchange Reactions

+ +

AD XZ AZ XD

AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)

Pb(NO3)2(aq) + K2CrO4(aq)

→ PbCrO4(s) + 2 KNO3(aq)

1. Write an unbalanced equation with correctformulas for all substances.

2. Balance the atoms of one of the elements.

Start with the most complex molecule.

Change the coefficients in front of the molecules.

Do NOT alter the chemical formulas.

3. Balance the remaining elements.

Balancing Chemical Equations

step 2 Al + Fe2O3 Al2O3 + Fe

1 Al (2Fe + 3O) (2Al + 3O) 2Fe

step 4 – balanced

22

Balance : Al + Fe2O3 Al2O3 + Fe

not balancednot balanced


balancedbalancednot balancednot balanced

step 3 Al + Fe2O3 Al2O3 + 2 Fe

2Al (2Fe + 3O) (2Al + 3O) 2Fe

22

step 1 Al + Fe2O3 Al2O3 + Fe1 Al (2Fe + 3O) (2Al + 3O) 1Fe


3


Balance C and N in C2H8N2 first:

C2H8N2 + N2O4 N2 + 44 H2O + 22 CO2

Combustion of rocket fuel:

C2H8N2 + N2O4 N2 + H2O + CO2

2C + 8H + 4N + 4O 1C + 2H + 2N + 3Onot balancednot balanced

2C + 8H + 4N + 4O 2C + 8H + 2N + 8OStill not balancednot balanced. Adjust N and O

C2H8N2 + 22 N2O4 33 N2 + 4 H2O + 2 CO2

BalancedBalanced

NaNO3(s) + H2SO4(aq) Na2SO4(aq) + HNO3(aq)


Balance Na in Na2SO4

Polyatomic ion on bothboth sides of an equation?• Balance as “units”.

NaNO3(s) + H2SO4(aq) Na2SO4(aq) + HNO3(aq)

Na + NO3 + 2H + SO4 2Na + SO4 + H + NO3


222Na + 2NO3 + 2H + SO4 2Na + SO4 + H + NO3


2 NaNO3(s) + H2SO4(aq) Na2SO4(aq) + 22 HNO3(aq)2Na + 2NO3 + 2H + SO4 2Na + SO4 + 2H + 2NO3

balancedbalanced

2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(ℓ)

The Mole and Chemical Reactions

Mole ratiosMole ratios:

2 mol C2H6

7 mol O2

=17 mol O2

2 mol C2H6=1

2 moles of C2H6 react with 7 moles of O2

2 moles of C2H6 produce 4 moles of CO2

2 mol C2H6 ≡ 7 mol O2

2 mol C2H6 ≡ 4 mol CO2 etc.

What mass of O2 and Br2 is produced by the reactionof 25.0 g of TiO2 with excess BrF3?

Notes:

• Check the equation is balanced!

• Stoichiometric ratios:

3TiO2 ≡ 3O2 ; 3TiO2 ≡ 2Br2 ; and many others

•• ExcessExcess BrF3 = enough BrF3 to react all the TiO2.

3 TiO2(s) + 4 BrF3(ℓ) → 3 TiF4(s) + 2 Br2(ℓ) + 3 O2(g)


= 25.0 g x = 0.3130 mol TiO2

1 mol79.88 g

nTiO2 = mass TiO2 / FM TiO2

What mass of O2 and Br2 is produced by the reaction of 25.0g of TiO2 withexcess BrF3? 3 TiO2(s) + 4 BrF3(ℓ) → 3 TiF4(s) + 2 Br2(ℓ) + 3 O2 (g)

3 mol TiO2 ≡ 3 mol O2

0.3130 mol TiO2 = 0.3130 mol O23 mol O2

3 mol TiO2


Mass of O2 produced = nO2 (mol. wt. O2)

= 0.3130 mol x 32.00 g/mol

= 10.0 g

3 TiO2 ≡ 2 Br2

nBr2 = 0.3130 mol TiO2 = 0.2087 mol Br2

2Br2

3 TiO2

Mass of Br2 = 0.2087 mol = 33.4 g Br2159.81 gmol Br2

What mass of O2 and Br2 is produced by the reaction of 25.0g of TiO2 withexcess BrF3? 3 TiO2(s) + 4 BrF3(ℓ) → 3 TiF4(s) + 2 Br2(ℓ) + 3 O2(g)


4

Practice Problem 4.8The purity of Mg can be found using the reaction…

Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g)

Calculate the % Mg in a 1.72-g sample that produced

6.46 g of MgCl2 when reacted with excess HCl.

More difficult – What should you calculate?

• How much purepure Mg will make 6.46 g of MgCl2?

• Express as a % of the original mass.

Practice Problem 4.8Mg +2 HCl → MgCl2 + H2

FW of MgCl2 = 24.31 + 2(35.45) = 95.21 g/mol

nMgCl2 = 6.46 g MgCl21 mol

95.21 g

Use mole ratioUse mole ratio 1 mol Mg ≡ 1 mol MgCl2

= 0.06785 mol MgCl2

Mg required: 0.06785 mol MgCl21 Mg

1 MgCl2

= 0.06785 mol of purepure Mg

0.06785 mol Mg x = 1.649 g Mg24.31 g1 mol

Practice Problem 4.8

Mg + 2 HCl → MgCl2 + H2

Calculate mass of pure Mg neededCalculate mass of pure Mg needed

1.649 g1.72 g

Given 1.72 g of impure Mg.

Purity (as mass %) = x 100% = 95.9 %

Reactions with Reactant in Limited Supply

Given 10 slices of cheese and 14 slicesof bread. How many sandwiches canyou make?

Balanced equation1 cheese + 2 bread 1 sandwich

1 cheese ≡ 2 bread1 cheese ≡ 1 sandwich2 bread ≡ 1 sandwich

Two methods can be used:

Product MethodProduct MethodCalculate the product from each starting material.

• The reactant giving the smallestsmallest number is limiting.

10 cheese x = 10 sandwiches1 sandwich1 cheese

14 bread x = 7 sandwiches1 sandwich2 bread

Correct answerBread is limiting. It will

be used up first


Reactant MethodReactant Method

Pick a reactant; calculate the amount of the other(s)needed. Enough?

e.g. choose breadcheese needed: 14 bread (1 cheese /2 bread ) = 7Available …. bread is limitingbread is limiting.

e.g. choose cheesebread needed: 10 cheese (2 bread/1 cheese) = 20Not available …. bread is limitingbread is limiting.

• Yes = Your choice is the limiting reactant.

• No = Another reactant is limiting.


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…base all other calculations on the limiting reactant.

Bread is limiting…Bread is limiting…

Sandwiches madeSandwiches made14 bread (1 sandwich / 2 bread ) = 7 sandwiches

Cheese remainingCheese remaining14 bread (1 cheese / 2 bread ) = 7 cheese used.Started with 10 cheese. Cheese remaining

10 – 7 = 3 slices


How much water will be produced by the combustionof 25.0 g of H2 in the presence of 100. g of O2?

Write a balanced equation:

nH2 = 25.0 g = 12.40 mol H2

1 mol H2

2.016 g

2 H2(g) + O2(g) 2 H2O(ℓ)

nO2 = 100. g = 3.125 mol O2

1 mol O2

32.00 g


2 H2 + O2 2 H2OMoles: 12.40 3.125

O2 gave less. OO22 is limitingis limiting. Use O2 in all calcs.

Product MethodProduct Method

From H2 nH2O = 12.40 mol H2 = 12.40 mol2H2O

2H2

From O2 nH2O = 3.125 mol O2 = 6.250 mol2H2O

1O2

mH2O = 6.250 mol H2O = 113. g water18.02 g

1 mol


2 H2 + O2 2 H2O

Moles available: 12.40 3.125

e.g. choose H2

O2 needed: 12.40 mol H2 (1 O2 /2 H2)= 6.20 molNot available …. OO22 is limitingis limiting.

You only need one calculation. Had you chosen O2

H2 needed: 3.125 mol O2 (2 H2 /1 O2)= 6.250 molAvailable …. OO22 is limitingis limiting.

H2O formed: 3.125 mol O2 (2H2O/1O2) = 6.250 mol.

= 113. g

Reactant MethodReactant Method


Consider :

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

If 374 g of NH3 and 768 g of O2 are mixed, what massof NO will form?

1 mol17.03 g

nNH3 = 374 g = 21.96 mol

1 mol32.00 g

nO2 = 768g = 24.00 mol

Balanced equation? yesyes

Reactions with Limited Reactants

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

Mol available: 21.96 24.00

From NHFrom NH33

NO formed: 21.96 mol NH3 = 21.96 mol NO

From OFrom O22

NO formed: 24.00 mol O2 = 19.20 mol NO

Smallest amount…. OO22 is limitingis limiting.

4 NO4 NH3

4 NO5 O2


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Mass of NO = 19.20 mol NO x = 576 g NO

OO22 is limitingis limiting. Base all calculationscalculations on O2.

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

Mass of NO formed?

30.01g1 mol

NO formed = 19.20 mol NO

21.96 mol 24.00 mol 19.20 mol

Reactions with Limited ReactantsWhat mass of MgI2 is made by the reaction of 75.0 gof Mg with 75.0 g of I2?

Mg + I2 → MgI2

• Balanced? YESYES

• Calculate moles

75.0 g of Mg = 75.0g/(24.31 g mol-1) = 3.085 mol Mg

75.0 g of I2 = 75.0g/(253.9 g mol-1) = 0.2955 mol I2

Limiting reactant? 1Mg ≡ 1I2 so II22 = limiting= limiting

• Since 1MgI2 ≡ 1I2 produce 0.2955 mol MgI2• Mass of MgI2 = 0.2955 mol x 278.2 g/mol = 82.2 g82.2 g


Theoretical yieldTheoretical yield

The amount of product predicted by stoichiometry.

Actual yieldActual yield

The quantity of desired product actually formed.

Percent yieldPercent yield

% yield = x 100%Actual yield

Theoretical yield

Percent Yield Percent Yield

Few reactions have 100% yield.

Possible reasons

Side reactionsSide reactions may produce undesired product(s).

Product lossProduct loss during isolation and purification.

Incomplete reactionIncomplete reaction due to poor mixing or reachingequilibrium…

Percent Yield

2.50 g of copper heated with an excess of sulfur made2.53 g of copper(I) sulfide

16 Cu(s) + S8(s) 8 Cu2S(s)

What was the percent yield for this reaction?

nCu used:= 0.03934 mol Cu

1 mol63.55g

2.50 g

16 mol Cu used 8 mol Cu2S made

Theoretical yieldTheoretical yield:

0.03934 mol Cu = 0.01967 mol Cu2S8 Cu2S16 Cu

Percent Yield2.50 g Cu + S8 (excesss) made 2.53 g Cu2S… What was the %-yield?

Theoretical yieldTheoretical yield = 0.01967 mol Cu2S

159.2 g1 mol

= 0.01967 mol Cu2S = 3.131 g Cu2S

Actual yieldActual yield = 2.53 g Cu2S (in problem)

Percent yieldPercent yield = x 100% = 80.8%2.53 g3.131 g

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Atom Economy

Examines the fate of all starting-material atoms.

High atom economy = low waste production

Ideal reaction: high % yield andand high atom economy.

atomic mass of atoms in useful product(s)atomic mass of all reactants used

x 100%

% atom economy =

Empirical formulaEmpirical formula = simplest ratio of atoms in amolecule.

Found for organic compounds by combustion analysiscombustion analysis

C and H are converted to CO2 & H2O.

• Both are trapped and the weight gain measured.

• Other elements (N, O …) with other traps or by massdifference.

H2Oabsorber

Mg(ClO4)2

CO2

absorber

NaOH

O2Sample ina furnace

Percent Composition & Empirical Formulas

Mass of CMass of C

1.502 g CO2 = 0.4099 g C1 mol CO2

44.009 g CO2

12.011 g C1 mol C

1 mol C1 mol CO2

oror

1.502 g CO2 = 0.4099 g C12.011 g C44.009 g CO2


Vitamin C (176.12 g/mol) contains C, H & O only. If1.000 g is burned in O2, 1.502 g CO2 and 0.409 g H2Oform. Find its empirical & molecular formula.

0.409 g H2O2.0158 g H

18.015 g H2O= 0.04577 g H

Mass of HMass of H

Vitamin C contains C, H & O only. Combustion of 1.000 g of vitamin C produced1.502 g of CO2 and 0.409 g of H2O.

mass of O = sample mass – (mass of C + mass of H)

Mass of OMass of O = 1.000 g – (0.4099 g + 0.04577 g)

= 0.544 g


= 0.04541 mol H0.04577 g H1.0079 g/mol

Convert to moles:

= 0.03413 mol C0.4099 g C12.011 g/mol

= 0.0340 mol O0. 544 g O

15.999 g/mol


Find the mole ratio (divide by smallest…):

C 0.03413 / 0.0340 = 1.00

H 0.04541 / 0.0340 = 1.34

O 0.0340 / 0.0340 = 1.00

Close to 1 : 1⅓ : 1 (C : H : O)

Multiply by 3 to get an integer ratio (3 : 4 : 3)

EmpiricalEmpirical formula is CC33HH44OO33


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Empirical mass = 3(12) + 4(1) + 3(16) g = 88 g

Molar mass = 176.12 g

Molar mass ≈ 2 x (empirical mass)

Vitamin C has the molecularmolecular formula C6H8O6

…Vitamin C contains C, H & O only… … molar mass of vitamin C is 176.12g/mol, find its empirical and molecular formula.

Empirical formula = C3H4O3


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