1

CHEMISTRYCHEMISTRYCHEMISTRY

Cof atom one of Mass X 121

element of atom one of Mass Amass,atomic lativeRe 12r

C of atom one of Mass x 121

molecule one of Mass

Mr mass, molecular Relative

12

abundance

)abundancemass (isotopic mass atomic Average

e% abundancmass) isotopicabundance (%

)2.93.05.90()u 222.9()u 213.0()u 205.90(

e% abundancmass) isotopicabundance (%

)91.3009.69()u 93.6491.30()u 93.6209.69(

abundancemassisotopic abundance

)85()u 025.1938()u 021.1915(

+ _ _

MATTER1.2 Mole Concept

Learning Outcome

At the end of this topic, students should be able to:

a) Define mole in terms of mass of carbon-12 and Avogadro’s constant, NA

Avogadro’s Number, NA

Atoms and molecules are so small – impossible to count

A unit called mole (abbreviated mol) is devised to count chemical substances by weighing them

A mole is the amount of matter that contains as many objects as the number of atoms in exactly 12.00 g of carbon-12 isotope

The number of atoms in 12 g of 12C is called Avogadro’s number, NA = 6.02 x 1023

Example:

1 mol of Cu contains Cu atoms

1 mol of O2 contains O2 molecules

O atoms

1 mol of NH3 contains NH3 molecules

N atoms

H atoms

6.02 6.02 10102323

33 6.02 6.02 10102323

22 6.02 6.02 101023236.02 6.02 10102323

6.02 6.02 10102323

6.02 6.02 10102323

1 mol of CuCl2 contains Cu2+ ions

Cl- ions

6.02 1023

2 6.02 1023

Mole and MassExample:

Relative atomic mass for carbon, C = 12.01

Mass of 1 C atom = 12.01 amu

Mass of 1 mol C atoms = 12.01 g

Mass of 1 mol C atoms consists of 6.02 x 1023 C atoms

= 12.01 g

Mass of 1 C atom =

= 1.995 x 10-23 g

2310 x 6.02 g 01.12

12.01 amu = 1.995 x 10-23 g

1 amu =

= 1.66 x 10-23 g

amu 12.01g 10 x .9951 23

Example:

From the periodic table, Ar of nitrogen, N is

� The mass of 1 N atom =

� The mass of 1 mol of N atoms =

� The molar mass of N atom =

� The molar mass of nitrogen gas =

The nucleon number of N =

14.0114.01

14.01 amu14.01 amu

14.01 g14.01 g

14.01 g mol14.01 g mol11

28.02 g mol28.02 g mol11

1414

Mr of CH4 is

� The mass of 1 CH4 molecule =

� The mass of 1 mol of CH4 molecules =

� The molar mass of CH4 molecule =

16.0516.05

16.05 amu16.05 amu

16.05 g16.05 g

16.05 g 16.05 g molmol11

Learning Outcome

At the end of this topic, students should be able to:(a) Interconvert between moles, mass,

number of particles, molar volume of gas at STP and room temperature.

(b) Define the terms empirical & molecular formulae

(c) Determine empirical and molecular formulae from mass composition or from mass composition or combustion data.combustion data.

Example 1:Calculate the number of moles of molecules for

3.011 x 1023 molecules of oxygen gas.

Solution:

6.02 x 1023 molecules of O2

3.011 x 1023 molecules of O2

= 0.5000 mol of O2 molecules

1 mol of O1 mol of O22 molecules molecules

molecules 10 6.02mol 1 molecules

23 2310011.3

Example 2:Calculate the number of moles of atoms for

1.204 x 1023 molecules of nitrogen gas.

Solution:

6.02 x 1023 molecules of N2

2 mol of N atoms

1.204 x 1023 molecules of N2

= 0.4000 mol of N atoms

1 mol of N1 mol of N22 molecules molecules

molecules mol 2 molecules

23

23

1002.610204.1

Example 3:Calculate the mass of 0.25 mol of chlorine gas.

Solution:

1 mol Cl2

0.25 mol Cl2

18 g

or

mass = mol x molar mass

= 0.25 mol x (2 x 35.45 g mol-1)

= 18 g

2 2 35.45 g 35.45 g

mol 1mol 0.25 g 35.45 2

Example 4:Calculate the mass of 7.528 x 1023 molecules of

methane, CH4Solution:

6.02 x 1023 CH4 molecules (12.01 + 4(1.01)) g

7.528 x 1023 CH4 molecules

= 20.06 g

23

23

1002.6107.528 g 05.16

Molar Volume of Gases Avogadro (1811) stated that equal volumes of gases

at the same temperature and pressure contain equal number of molecules

Molar volume is a volume occupied by 1 mol of gas

At standard temperature and pressure (STP), the molar volume of an ideal gas is 22.4 L mol 1

Standard Temperature and Pressure

273.15 K 1 atm 760 mmHg

0 C 101325 N m-2

101325 Pa

Standard Molar Volume

At room conditions (1 atm, 25 C), the molar volume of a gas = 24 L mol-1

Example 1:

Calculate the volume occupied by 1.60 mol of Calculate the volume occupied by 1.60 mol of ClCl22 gas at STP. gas at STP.Solution:At STP,At STP,1 mol Cl1 mol Cl22 occupiesoccupies

1.60 mol Cl1.60 mol Cl22occupies occupies

= 35.8 L= 35.8 L

22.4 L

mol 1L 4.22mol 60.1

Example 2:

Calculate the volume occupied by 19.61 g of Calculate the volume occupied by 19.61 g of NN22 at STP at STPSolution:1 mol of N1 mol of N22 occupiesoccupies 22.4 L22.4 L

of N of N22 occupies occupies

= 15.7 L= 15.7 L

mol 1

L 22.4mol )01.14(2

61.19

1mol g 2(14.01)g 61.19

Example 3:0.50 mol methane, CH0.50 mol methane, CH44 gas is kept in a cylinder at gas is kept in a cylinder at STP. Calculate:STP. Calculate:(a)(a) The mass of the gasThe mass of the gas(b)(b) The volume of the cylinderThe volume of the cylinder(c)(c) The number of hydrogen atoms in the cylinderThe number of hydrogen atoms in the cylinderSolution:(a)(a) Mass of 1 mol CHMass of 1 mol CH44 ==

Mass of 0.50 mol CHMass of 0.50 mol CH44 ==

= 8.0 g= 8.0 g

16.05 g

mol 1mol 0.50g 05.16

(b)(b)At STP;At STP; 1 mol CH1 mol CH44 gas gas occupiesoccupies

0.50 mol CH0.50 mol CH44 gas gas occupies occupies

= 11 L = 11 L

(c)(c) 1 mol of CH1 mol of CH44 molecules molecules 4 mol of H atoms4 mol of H atoms

0.50 mol of CH0.50 mol of CH44 molecules molecules 2 mol of H atoms2 mol of H atoms

1 mol of H atoms1 mol of H atoms

2 mol of H atoms2 mol of H atoms

1.2 x 10 1.2 x 102424 atoms atoms

22.4 L

6.02 x 1023 atoms

2 x 6.02 x 1023 atoms

mol 1mol 0.50 L 4.22

Exercise A sample of CO2 has a volume of 56 cm3 at STP.

Calculate:

a) The number of moles of gas molecules (0.0025 mol)

b) The number of CO2 molecules (1.506 x 1021 molecules)

c) The number of oxygen atoms in the sample (3.011x1021atoms)

Notes: 1 dm3 = 1000 cm3

1 dm3 = 1 L

Empirical And Molecular Formulae

Empirical formula => => chemical formula that shows the simplest ratio of all elements in a molecule.

Molecular formula => => formula that show the actual number of atoms of each element in a molecule.

The relationship between empirical formula and molecular formula is :

Molecular formula = n ( empirical formula )

Example:

A sample of hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its molar mass is 56. Determine the empirical formula and molecular formula of the compound.

== = = == = =

Mass

Number of moles

Simplest ratio

85.7

21.984

14.1584

1

7.1357

14.3

01.127.85

01.13.14

C H

Empirical formula = Empirical formula = CH 2

14.0356 n

4 3.99

84

2

HC formula Molecular )n(CH formula Molecular

Exercise: A combustion of 0.202 g of an organic sample that contains carbon, hydrogen and oxygen produce 0.361g carbon dioxide and 0.147 g water. If the relative molecular mass of the sample is 148, what is the molecular formula of the sample?

Answer : C6H12O4

At the end of this topic, students should be At the end of this topic, students should be able to:(a) (a) Define and perform calculation for each of for each of

the the following concentration measurements :following concentration measurements : i) molarity (M) ii) molality(m) iii) mole fraction, X iv) percentage by mass, % w/w v) percentage by volume, %v/v

Learning Outcome

Concentration of Solutions

A solution is a homogeneous mixture of two or more substances:

solvent + solute(s)

e.g: sugar + water – solution

sugar – solute

water – solvent

Concentration of a solution can be expressed in various ways :

a) molarityb) molalityc) mole fractiond) percentage by mass e) percentage by volume

a) Molarity

Molarity is the number of moles of solute in 1 litre of solution

Units of molarity: mol L-1

mol dm-3

M

(L) solution of volume(mol) solute of molesM molarity,

Example 1: Determine the molarity of a solution

containing 29.22 g of sodium chloride, NaCl in a 2.00 L solution.

Solution:

solution

NaClNaCl V

n M

L 00.2

mol )45.3599.22(

22.29

= 0.250 mol L-1

Example 2:How many grams of calcium chloride, CaCl2 should be used to prepare 250.00 mL solution with a concentration of 0.500 M

Solution:

solutionCaClCaCl x VM n22

= 0.500 mol L 1 250.00 10 3 L

massmolar x n CaClof mass2CaCl2

= (0.500 250.00 10 3 ) mol (40.08 + 2(35.45)) g mol 1

= 13.9 g

b) Molality

Molality is the number of moles of solute dissolved in 1 kg of solvent

Units of molality:mol kg-1

molal m

(kg) solvent of mass(mol) solute of moles m molality,

Example:What is the molality of a solution prepared by dissolving 32.0 g of CaCl2 in 271 g of water?

Solution:

1-CaCl mol g 2(35.45) 08.40g 0.32

n2

kg 10271

mol 98.1100.32

CaClof Molality 32

1kg mol 1.06

Exercise:Calculate the molality of a solution prepared by dissolving 24.52 g of sulphuric acid in 200.00 mL of distilled water. (Density of water = 1 g mL-1)

Ans = 1.250 mol kg-1

c) Mole Fraction (X)

Mole fraction is the ratio of number of moles of one component to the total number of moles of all component present.

For a solution containing A, B and C:

T

A

CBA

AA

nn

n n nn

Xof A, fraction Mol

Mol fraction is always smaller than 1

The total mol fraction in a mixture (solution) is equal to one.

XA + XB + XC + X….. = 1

Mole fraction has Mole fraction has no unit (dimensionless) since it is a ratio of two similar quantities.

Example:A sample of ethanol, C2H5OH contains 200.0 g of ethanol and 150.0 g of water. Calculate the mole fraction of(a) ethanol(b) waterin the solution.

Solution:

n ethanol = 1mol g 16.00)5(1.01) (2(12.01)g 0.200

n water = 1mol g 16.00)(2(1.01)g 0.150

X ethanol =

mol 02.18

0.150mol

07.450.200

mol 07.45

0.200

= 0.3477

X water = 1 0.3477= 0.6523

d) Percentage by Mass (%w/w)

Percentage by mass is defined as the percentage of the mass of solute per mass of solution.

Note: Mass of solution = mass of solute + mass of solvent

100xsolutionof masssoluteof mass

ww%

Example:A sample of 0.892 g of potassium chloride, KCl is dissolved in 54.362 g of water. What is the percent by mass of KCl in the solution?

Solution:

%100g 54.362g 0.892

g 892.0 mass %

= 1.61%

Exercise:A solution is made by dissolving 4.2 g of sodium chloride, NaCl in 100.00 mL of water. Calculate the mass percent of sodium chloride in the solution.

Answer = 4.0%

e) Percentage by Volume (%V / V)

Percentage by volume is defined as the percentage of volume of solute in milliliter per volume of solution in milliliter.

Note:

solutionof volume solutionof mass

solutionof Density

100 x solution of volumesolute of volume v

v % 100 x solution of volumesolute of volume v

v %

Example 1: 25 mL of benzene is mixed with 125 mL

of acetone. Calculate the volume percent of benzene solution.

Solution:

100% mL 125 mL 25

mL 25 volume%

= 17%

Example 2:A sample of 250.00 mL ethanol is labeled as 35.5% (v/v) ethanol. How many milliliters of ethanol does the solution contain?

Solution:

100%mL 00.250%5.35

Vethanol

%100VV

ethanolof volume%solution

ethanol

= 88.8 mL

Example 3:A 6.25 m of sodium hydroxide, NaOH solution has has a density of 1.33 g mL-1 at 20 ºC. Calculate the concentration NaOH in:(a) molarity(b) mole fraction(c) percent by mass

(a) M = solution

NaOHVn

6.25 m of NaOH there is 6.25 mol of NaOH in 1 kg of water

for a solution consists of 6.25 mol of NaOH and 1 kg of water;

V solution = solution

solutionmass

mass solution = mass NaOH + mass water

mass NaOH = n NaOH molar mass of NaOH = 6.25 mol (22.99 + 16.00 + 1.01) g mol 1

= 250 g

mass solution = 250 g + 1000 g = 1250 g

V solution = 1mL g 1.33g 1250

M NaOH =

L 10

33.11250

mol 25.63

= 6.65 mol L 1

(b) X NaOH = waterNaOH

NaOHnn

n

1 kg of water contains 6.25 mol of NaOH

n water = waterof mass molar

mass water

= 1mol g 16.00) (2(1.01)g 1000

X NaOH =

mol

02.181000 mol 25.6

mol 25.6

= 0.101

(c) %(w/w) of NaOH = waterNaOH

NaOHmassmass

mass

100%

= g 1000 g 250

g 250

100%

= 20.0%

Exercise:An 8.00%(w/w) aqueous solution of ammonia has a density of 0.9651 g mL-1. Calculate the(a) molality (b) molarity(c) mole fraction

of the NH solution

Answer: a) 5.10 mol kg-1

b) 4.53 mol L-1

c) 0.0842

MATTER1.3 Stoichiometry

Learning Outcome

At the end of the lesson, students should be able to:

a) Determine the oxidation number of an element in a chemical formula.

b) Write and balance :i) Chemical equation by inspection methodii) redox equation by ion-electron method

Balancing Chemical Equation

A chemical equation shows a chemical reaction using symbols for the reactants and products.

The formulae of the reactants are written on the left side of the equation while the products are on the right.

Example:

x A + y B z C + w D

Reactants Products

A chemical equation must have an equal number of atoms of each element on each side of the arrow

The number x, y, z and w, showing the relative number of molecules reacting, are called the stoichiometric coefficients.

A balanced equation should contain the smallest possible whole-number coefficients

The methods to balance an equation: a) Inspection method b) Ion-electron method

Inspection Method

1. Write down the unbalanced equation. Write the correct formulae for the reactants and products.

2. Balance the metallic atom, followed by non-metallic atoms.

3. Balance the hydrogen and oxygen atoms.

4. Check to ensure that the total number of atoms of each element is the same on both sides of equation.

Example: Balance the chemical equation by applying the inspection method.

NH3 + CuO → Cu + N2 + H2O

Exercise

Balance the chemical equation below by applying inspection method.

1. Fe(OH)3 + H2SO4 → Fe2(SO4)3 + H2O

2. C6H6 + O2 → CO2 + H2O

3. N2H4 + H2O2 → HNO3 + H2O

4. ClO2 + H2O → HClO3 + HCl

Redox Reaction

Mainly for redox (reduction-oxidation) reaction

Oxidation is defined as a process of electron loss. The substance undergoes oxidation

loses one or more electrons.

increase in oxidation number

act as an reducing agent (electron donor)

Half equation representing oxidation: Mg Mg2+ 2e Fe2+ Fe3+ + e 2Cl- Cl2 + 2e

Reduction is defined as a process of electron gain. The substance undergoes reduction

gains one or more electrons.

decrease in oxidation number

act as an oxidizing agent (electron acceptor)

Half equation representing reduction: Br2 + 2e → Br-

Sn4+ + 2e → Sn2+

Al3+ + 3e → Al

Oxidation numbers of any atoms can be determined by applying the following rules:

1. For monoatomic ions, oxidation number = the charge on the ione.g: ion oxidation number

Na+ +1Cl- -1Al3+ +3S2- -2

2. For free elements, e.g: Na, Fe, O2, Br2, P4, S8 oxidation number on each atom = 0

1. For most cases, oxidation number for O = -2 H = +1

Halogens = -1

Exception:

1. H bonded to metal (e.g: NaH, MgH2) oxidation number for H = -1

2. Halogen bonded to oxygen (e.g: Cl2O7) oxidation number for halogen = +ve

3. In a neutral compound (e.g: H2O, KMnO4) the total of oxidation number of every atoms that made up the molecule = 0

4. In a polyatomic ion (e.g: MnO4-, NO3

-) the total oxidation number of every atoms that made up the molecule = net charge on the ion

Exercise

1. Assign the oxidation number of Mn in the following chemical compounds.i. MnO2 ii. MnO4

-

1. Assign the oxidation number of Cl in the following chemical compounds.i. KClO3 ii. Cl2O7

2-

1. Assign the oxidation number of following:i. Cr in K2Cr2O7ii. U in UO2

2+

iii. C in C2O42-

Balancing Redox Reaction

Redox reaction may occur in acidic and basic solutions.

Follow the steps systematically so that equations become easier to balance.

Balancing Redox Reaction In Acidic Solution

Fe2+ + MnO4- → Fe3+ + Mn2+

1. Separate the equation into two half-reactions: reduction reaction and oxidation reactioni. Fe2+ → Fe3+ ii. MnO4

- → Mn2+

1. Balance atoms other than O and H in each half-reaction separately

i. Fe2+ → Fe3+

ii. MnO4- → Mn2+

3. Add H2O to balance the O atomsAdd H+ to balance the H atoms

i. Fe2+ → Fe3+ ii. MnO4

- + → Mn2+ +

4. Add electrons to balance the charges

i. Fe2+ → Fe3+ + ii. MnO4

- + 8H+ + → Mn2+ + 4H2O

4H2O8H+

1 e

5 e

3.Multiply each half-reaction by an integer, so that number of electron lost in one half-reaction equals the number gained in the other.

i. 5 x (Fe2+ → Fe3+ + 1e)5Fe2+ → 5Fe3+ + 5e

ii. MnO4- + 8H+ + 5e → Mn2+ + 4H2O

1. Add the two half-reactions and simplify where possible by canceling the species appearing on both sides of the equation.

i. 5Fe2+ → 5Fe3+ + 5eii. MnO4

- + 8H+ + 5e → Mn2+ + 4H2O___________________________________5Fe2+ + MnO4

- + 8H+ → 5Fe3+ + Mn2+ + 4H2O

___________________________________

5. Check the equation to make sure that there are the same number of atoms of each kind and the same total charge on both sides.

5Fe 2+ + MnO 4- + 8H + → 5Fe 3+ + Mn 2+ + 4H 2O

Total charge reactant= 5(+2) + ( -1) + 8(+1)= + 10 - 1 + 8= +17

Total charge product= 5(+3) + (+2) + 4(0)= + 15 + (+2)= +17

5Fe 2+ + MnO 4- + 8H + → 5Fe 3+ + Mn 2+ + 4H 2O

Total charge reactant= 5(+2) + ( -1) + 8(+1)= + 10 - 1 + 8= +17

Total charge product= 5(+3) + (+2) + 4(0)= + 15 + (+2)= +17

Exercise: In Acidic Solution

C2O42- + MnO4

- + H+ → CO2 + Mn2+ + H2O

Solution :

Balancing Redox Reaction In Basic Solution

1. Firstly balance the equation as in acidic solution.

2. Then, add OH- to both sides of the equation so that it can be combined with H+ to form H2O.

3. The number of OH- added is equal to the number of H+ in the equation.

Example: In Basic Solution

Cr(OH)3 + IO3- + OH- → CrO3

2- + I- + H2O

Exercise:

1. H2O2 + MnO4- + H+ O2 + Mn2+ + H2O

(acidic medium)2. Zn + SO4

2- + H2O Zn2+ + SO2 + 4OH-

(basic medium)3. MnO4

- + C2O42- + H+ Mn2+ + CO2 + H2O

(acidic medium)4. Cl2 ClO3

- + Cl- (basic medium)

Stoichiometry

Stoichiometry is the quantitative study of reactants and products in a chemical reaction.

A chemical equation can be interpreted in terms of molecules, moles, mass or even volume.

C3H8 + 5O2 3CO2 + 4H2O

1 molecule of C3H8 reacts with 5 molecules of O2 to produce 3 molecules of CO2 and 4 molecules of H2O

6.02 x 1023 molecules of C3H8 reacts with 5(6.02 x 1023) molecules of O2 to produce 3(6.02 x 1023) molecules of CO2 and 4(6.02 x 1023) molecules of H2O

C3H8 + 5O2 3CO2 + 4H2O

1 mol of C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O

44.09 g of C3H8 reacts with 160.00 g of O2 to produce 132.03 g of CO2 and 72.06 g of H2O

5 moles of C3H8 reacts with 25 moles of O2 to produce 15 moles of CO2 and 20 moles of H2O

At room condition, 25 ºC and 1 atm pressure;22.4 dm3 of C3H8 reacts with 5(22.4 dm3) of O2 to produce 3(22.4 dm3) of CO2

Example 1:How many grams of water are produced in the oxidation of 0.125 mol of glucose?

C6H12O6(s) + O2(g) CO2(g) + H2O(l)

Solution:From the balanced equation;1 mol C6H12O6 produce 6 mol H2O

0.125 mol C6H12O6 produce H2O

mass of H2O = (0.125 x 6) mol x (2.02 + 16.00) g mol-1

= 13.5 g

mol 1mol 6mol 125.0

Example 2:Ethene, C2H4 burns in excess oxygen to form carbon dioxide gas and water vapour.(a) Write a balance equation of the

reaction(b) If 20.0 dm3 of carbon dioxide gas is

produced in the reaction at STP, how many grams of ethene are used?

Solution:(a) C2H4 + O2 CO2 + H2O

(b) 22.4 dm3 is the volume of 1 mol CO2

20.0 dm3 is the volume of CO2

2 mol CO2 produced by 1 mol C2H4

mol CO2 produced by C2H4

3

3

dm 4.22mol 1dm 0.20

4.220.20

mol 2

mol 1mol 4.220.20

mass ethane1-mol g 4(1.01)] [2(12.01) x mol

24.220.20

= 12.5 g

Learning Outcome

At the end of this topic, students should be able to:

a) Define the limiting reactant and percentage yield

b) Perfome stoichiometric calculations using mole concept including limiting reactant and percentage yield.

Limiting Reactant/Reagent

Limiting reactant is the reactant that is completely consumed in a reaction and limits the amount of product formed

Excess reactant is the reactant present in quantity greater than necessary to react with the quantity of limiting reactant

Example:3H2 + N2 2NH3If 6 moles of hydrogen is mixed with 6 moles of nitrogen,

how many moles of ammonia will be produced?

Solution:3 mol H2 reacts with 1 mol N2

6 mol H2 reacts with

mol 3

mol 1 mol 6

= 2 mol N 2

N2 is the excess reactantH2 is the limiting reactant limits the amount of products formed

3 mol H2 produce 2 mol NH3

6 mol H2 produce

mol 3mol 2 mol 6

= 4 mol NH 3

or1 mol N2 react with 3 mol H2

6 mol N2 react with mol NH3

mol 1mol 3 mol 6

= 18 mol H 2

H 2 is not enough

H 2 limits the amount of products formed

limiting reactant

3 mol H2 produce 2 mol NH3

6 mol N2 produce mol NH3

= 4 mol NH3

mol 3mol 2 mol 6

Exercise:Consider the reaction:

2 Al(s) + 3Cl2(g) 2 AlCl3(s)

A mixture of 2.75 moles of Al and 5.00 moles of Cl2 are allowed to react.

(a) What is the limiting reactant?(b) How many moles of AlCl3 are formed?(c) How many moles of the reactant remain at the end of the reaction?

PERCENTAGE YIELD

The amount of product predicted by a balanced equation is the theoretical yield

The theoretical yield is never obtain because: 1. The reaction may undergo side reaction 2. Many reaction are reversible 3. There may be impurities in the reactants

4. The product formed may react further to form other product

5. It may be difficult to recover all of the product from the reaction medium

The amount product actually obtained in a reaction is the actual yield

Percentage yield is the percent of the actual yield of a product to its theoretical yield

100 x yield

yield yield %ltheoretica

actual

Example 1:Benzene, C6H6 and bromine undergo reaction as follows:C6H6 + Br2 C6H5Br + HBr

In an experiment, 15.0 g of benzene are mixed with excess bromine

(a) Calculate the mass of bromobenzene, C6H5Br that would be produced in the reaction.

(b) What is the percent yield if only 28.5 g of bromobenzene obtain from the experiment?