Chemistry Ch 1

F.S.C part1 Chap 1st

Posted by chismtry zone on Sunday, 18 December 2011 at 01:33

Chapter No. 1

BASIC CONCEPTS

MCQs

Q.1 Smallest particle of an element which may or may not have independent existence (a) a molecule (b) an atom

(c) an ion (d) an electron Q.2 Swedish chemist J. Berzelius determined the (a) atomic no. (b) atomic volume (c) atomic mass (d) atomic density Q.3 The number of atoms present in a molecule determine its (a) molecularity (b) basicity

(c) acidity (d) atomicity

Q.4 When an electron is added to a unipositive ion we get (a) anion (b) cation

(c) neutral atom (d) molecule Q.5 CO

+

is an example of:

(a) free radical (b) cationic molecular ion (c) an ionic molecular ion

(d) stable molecule

Q.6 Relative atomic mass is the mass of an atom of an element as compared to the mass of (a) oxygen (b) hydrogen

(c) nitrogen (d) carbon

Q.7 Isotopes are the sister atoms of the same element with similar chemical properties and different (a) atomic number (b) atomic mass

Q.8 The instrument which is used to measure the exact masses of different isotopes of an element called

(a) I.R. Spectrophotometer (b) U.V. Spectrophotometer (c) Mass Spectrometer (d) Colourimeter

Q.9 Mass spectrometer separates different positive isotopic ions on the basis of their (a) mass value (b) m/e value

(c) e/m value (d) change value

Q.10 Simplest formula that gives us information about the simple ratio of atoms in a compound is called

(a) structural formula (b) molecular formula (c) empirical formula (d) molar ratio

Q.11 Percentage of oxygen in H

2

O is

(a) 80% (b) 88.8% (c) 8.8% (d) 9.8%

Q.12 More abundant isotope of an element is one with

(a) even atomic no. (b) odd atomic no. (c) Even mass no. (d) odd mass no.

Q.13 Large no. of isotopes are known for the elements whose masses are multiple of (a) two (b) four

(c) six (d) eight

Q.14 When 0.01 kg of CaCO

3

is decomposed the CO

2

produced occupies a volume at S.T.P. (a) 2.2414 dm

3

(b) 22.414 dm

3

(c) 22414 dm

3

(d) 224014 dm

3

Q.15 The no. of covalent bond in 10gm of NH

3

are

(a) 6.022 x 10

23

(b) 1.062 x 10

23

(c) 10.62 x 10

24

(d) 1.062 x 10

24

Q.16 No. of molecules present in 10gm of water are (a) 3.37 x 10

23

(b) 33.7 x 10

23

(c) 3.37 x 10

24

(d) 3.037 x 10

24

Q.17 The no. of covalent bonds present in 10gm of water are (a) 6.074 x 10

23

(b) 6.74 x 10

23

(c) 6.074 x 10

24

(d) 6.74 x 10

24

Q.18 The least no. of molecules present in 30 gm of (a) N

2

O (b) NO

(c) NO

2

(d) N

2

O

3

Q.19 Which of the following has highest percentage of nitrogen (a) (NH

4

)

2

SO

4

(b) NH

4

H

2

PO

4

(c) (NH

4

)

2

HPO

4

(d) (NH

4

)

3

PO

4

Q.20 0.1 mole of Na

3

PO

4

completely dissociates in water to produce Na

+

(a) 6.02 x 10

22

(b) 6.02 x 10

23

(c) 1.806 x 10

23

(d) 1.806 x 10

22

Q.21 Efficiency of chemical reaction can be checked by calculating (a) amount of limiting reactant

(b) amount of the reactant in excess (c) amount of the product formed (d) amount of the reactant unused Q.22 A limiting reactant is one

(a) which is present in least amount

(b) which produces minimum no. of moles of product (c) which produces maximum no. of moles of product (d) does not effect the amount of product

Q.23 Stoichiometry is the branch of chemistry which deals with the study of quantitative relationship among the various

(a) reactants (b) products

(c) Reactants and products (d) all of above Q.24 500 cm

3

of H

2

gas at STP contradictions of hydrogen (a) 6.02 x 10

23

(b) 3.01 x 10

22

(c) 2.68 x 10

22

(d) 1.34 x 10

22

Q.25 Largest number of H

+

ions are produced by complete ionization of (a) 0.01 mole of HCl (b) 0.0050 mole of H

2

SO

4

(c) 0.000334 moles of H

3

PO

4

(d) all above

Q.26 The Avogadro’s number is

(a) 6.02 x 10

24

(b) 6.02 x 10

–24

(c) 6.02 x 10

–23

(d) 6.02 x 10

23

Q.27 The largest number of H

+

are produced by complete ionization of (a) 0.100 2 moles of HCl (b) 0.051 moles of H

2

SO

4

(c) 0.0334 moles of H

3

PO

4

(d) All of the above Q.28 A sample of pure matter is

(a) element (b) compound (c) substance (d) mixture

Q.29 nm stands for

(a) Newton meter (b) Nanometer

(c) Newton square meter (d) none of the above Q.30 One calorie is equal to

(a) 4.184 J (b) 41.84 J (c) 0.4184 J (d) 0.04184 J

Q.31 The number of moles of CO

2

which contains 8.0 gm of oxygen (a) 0.25 (b) 0.50

(c) 1.0 (d) 1.50

Q.32 27 grams of Al will react completely with how much mass of O

2

to produce Al

2

O

3

(a) 8 gm of oxygen (b) 16 gm of oxygen

(c) 32 gm of oxygen (d) 24 gm of oxygen Q.33 Mole of SO

2

contains

(b) 18.1 x 10

23

molecules of SO

2

(c) 6.023 x 10

23

atom of sulphur (d) 4 gram of SO

2

Q.34 The largest number of molecules are presenting

(a) 3.6 gram of H

2

O (b) 4.8 gram of C

2

H

5

OH (c) 2.8 gm of CO (d) 5.4 gms of N

2

O

5

Q.35 The mass of one mole of electron is

(a) 1.008 mg (b) 0.184 mg (c) 1.673 mg (d) 0.55 mg Q.36 Isotopes differ in

(a) properties which depend on mass (b) arrangements of electrons in orbital (c) chemical properties

(d) the extent to which they may be affected in electromagnetic field Q.37 The volume occupied by 1.4 gm of N

2

at STP is

(a) 224 dm

3

(b) 22.4 dm

3

(c) 1.12 dm

3

(d) 112 cm

3

Q.38 Many elements have fractional atomic mass. This is because (a) the mass atom is itself fractional

(b) atomic masses are average masses of isobars (c) atomic masses are averages masses of isotopes

(d) atomic masses are average masses of isotopes proportional to relative abundance Q.39 A limiting reactant is one which

(a) is taken in lesser quantity in grams as compared to other reactants (b) is taken in lesser quantity in volume as compared to the other (c) gives the maximum amount of the product which is required

(d) gives the minimum amount of the product under consideration Q.40 Isotopes when even atomic masses are a comparatively abundant (a) demper’s spectrograph is superior to that of Aston’s

(b) 0.1 mg of H

2

O has greater number of molecules then 0.1 mg of CH

4

(c) the number of H

+

and PO

–3

ions are not equal but the number of positive and negative charges

(d) are equal when 100 molecules of H

3

PO

4

are thrown in excess of water Q.41 A molecule having two atoms is called

(a) monoatomic molecules (b) diatomic molecules (c) Polyatomic molecules (d) homoatomic molecule Q.42 An ordinary misoscope is used to measure the object of size (a) upto 500 nm (b) upto 850 nm

(c) upto 1000 nm (d) upto 1200 nm Q.43 1 atomic masses unit (amu) is equation

(a) 1.66 x 10

–27

kg (b) 1.56 x 10

–27

kg (c) 1.76 x 10

–21

kg (d) 1.8 x 10

–27

kg Q.44 Nickel has isotopes

(a) 1 (b) 3 (c) 5 (d) 7 Q.45 Cadmium has isotopes

(a) 3 (b) 5 (c) 7 (d) 9

Q.46 The pressure of vapours in the separating isotopes by mass spectrometry is kept at (a) 10

–6

torr (b) 10

–4

torr

(c) 10

–3

torr (d) 10

–5

torr Q.47 Number of gram atoms in 0.1 gm of Na is (a) 0.0043 (b) 0.0403 (c) 0.403 (d) None of these Q.48 Molecule of haemoglobin contains atoms

(a) 15,000 (b) 12,000 (c) 10,000 (d) 8,000 Q.49 Haemoglobin is heavier than a hydrogen atom (a) 65,000 (b) 68,000 (c) 62,000 (d) 60,000 Answers

Questions

1

2

3

4

5

Answers

b

C

d

c

b

Questions

6

7

8

9

10

Answers

d

b

c

b

c

Questions

11

12

13

14

15

Answers

b

c

b

a

d

Questions

16

17

18

19

20

Answers

a

b

d

d

c

Questions

21

22

23

24

25

Answers

c

b

d

c

d

Questions

26

27

28

29

30

Answers

d

d

a

b

a

Questions

31

32

33

34

35

Answers

a

d

c

a

d

Questions

36

37

38

39

40

Answers

a

c

d

d

c

Questions

41

42

43

44

Answers

c

a

a

c

Questions

45

46

47

48

49

Answers

d

a

a

c

b

Short Question With Answer

Q.1 Calculate the grams atoms in 0.4 gm of potassium.

Ans.

Gram atoms of potassium =

= = 0.01 grams atoms

Q.2 23 grams of sodium and 238 gram of uranium have equal number of atoms in them.

Ans.

Mass of sodium = 23 gms= 1mole=6.02 x 10

23

atoms Mass of uranium = 238g=1 mole= 6.02x 10

23

atoms

Both the substances have equal number of atoms because they have same no.of moles.

Q.3 Mg atom is twice heavier than that of carbon.

Ans.

The atomic mass of Mg is 24 which is to twice as mass as compared to the atomic mass of carbon i.e. 12. So Mg atom is twice heavier than that of carbon.

Q.4 180 grams of glucose and 342 gram of sucrose have the same number of molecules but different

number of atoms present in them.

Ans.

180 grams of glucose (C

6

H

12

O

6

) and 342 grams of sucrose (C

12

H

22

O

11

) are their molar masses indicating one mole of each (glucose and sucrose) one mole of a substance contains equal number of molecules i.e. 6.02 x 10

23

.

Mass of glucose (C6H12O6)=180g=1mole=6.02x10

23

molecules

=24NAatoms

Mass of Sucrose (C12H22O11)=342g=1mole=6.02 x 10

23

molecules

=45NAatoms

Q.5 4.9 g of H

2

SO

4

when completely ionized in water have equal number of positive and negative ions, but the number of positively charged ions are twice the number of negatively charged ions.

Ans.

H2SO4 → 2H +

+ SO4 -2

When one mole of H

2

SO

4

ionizes, it produces 2H

+

and

SO

4–2

ions. Hydrogen ions contains +1 charge while sulphate ions have – 2 charge. Hydrogen ions are twice in number than that of SO ion. Charges on both ions are equal (with opposite sign). Similarly ions produced by complete ionization of 4.9 grams of H

2

SO

4

in water will have equal +ve and –ve charges but the number of H

+

ions are twice than number of negatively charged sulphate ions.

Q.6 One mg of K

2

CrO

4

has thrice the number of ions than the number of molecules when ionized in excess of water.

Ans. K2CrO4 → 2K +

+ CrO

4–2

When K

2

CrO

4

ionizes in water, its one molecule gives three ions i.e. two K

+

and one CrO

4–2

(chromate) ions. The ratio between the number of molecules and number of ions than the number of molecules when ionized in water.

Q.7 Two grams of H

2

, 16g of CH

4

and 44 gram of CO

2

occupy separately the volumes of 22.414 dm

3

at STP, although the sizes and masses of molecules of three gases are very different from each other.

Ans.

One mole of gas at STP occupies a volume of 22.4 dm

3

sizes and masses of molecules of different gas do not affect the volume. Normally it is known that in the gaseous state, the distance between the molecules is 300

times greater that their diameter. Therefore two grams of H

2

, 16 grams of CH

4

and 44 grams of CO

2

(1 mole of each gas) separately occupy a volume of 22.4 dm

3

. This is called molar volume.

2gH2=1mole, 16gCH4=1 mole, 44gCO2=1 mole

1mole=22.414dm3

Q.8 Define Stoichiometry ?

Ans.

Stoichiometry is the branch of chemistry which gives a quantitative relationship between reactants and products in balanced chemical equation.

Q.9 What is limiting reactant? How does it control the quantity of the product formed? Explain with

three examples. /Many chemical reactions taking place in our surroundings involve limiting reactants give examples?

Ans.

The reactant which controls (limits) the amount of product formed during a chemical reaction is called limiting reactant. In our surrounding many chemical reactions take place which involve limiting reactants

some of these reactions are:

(i) Burning of coal to form CO

2---Coal is limiting reactatnt

C + O2 CO2

(ii) Burning of sui gas to form CO

2

and H

2

O

CH

4

+ 2O

2

 CO

2

+ 2H

2

O

(iii) Rusting of iron----iron is limiting reactant

In above reactions oxygen is always in excess, while other reactants are consumed earlier. So other reactants are limiting reactants.

Q.10 One mole of H

2

O has two moles of bands, three moles of atoms, ten moles of electron and twenty–eight moles the total fundamental particles present in it.

Ans.

One molecule of H–O–H has two bounds between hydrogen and oxygen. There are three atoms i.e. two

H atoms and one O atom, therefore one mole of H

2

O contains two moles of bonds and three moles of atoms (2 moles of H atoms and one mole of O atoms).

Similarly, there are eight elections in oxygen and one electron in each of the two, H atoms one molecule of H

2

O so has 10 electrons, so one mole of water contains 10 moles of electrons. There are 28 moles of all fundamental particles in one mole of water i.e.

10 moles of electrons.

10 moles of protons.

8 moles of neurons (8 neutrons in oxygen and there is no neutron in hydrogen) 28 moles of fundamental particles.

Q.11 One mole of H

2

SO

4

should completely react with two moles of NaOH. How does Avogadro’s

number help to explain it?

Ans.

The balanced chemical equation between H

2

SO

4

and NaOH

H

2

SO

4

+ 2NaOH  Na

2

SO

4

+ H

2

O H

2

SO

4

 2H+ + SO4

-2

2NaOH2Na+ + 2OH 2H+ + 2OH-  2H2O

This is an acid base reaction, one mole of H

2

SO

4

releases two moles of H

+

ion in solution. It needs two moles of OH ions for complete neutralization. So two moles of NaOH which releases two moles of OH are required to react with one mole of H

2

SO

4

. One mole of H

2

SO

4

releases twice the Avogadro’s number of H

+

ions and it will need the Avogadro’s number of OH ions for complete neutralization.

Q.12 N

2

and CO have same number of electrons, protons and neutrons.

Ans.

Both N

2

and CO have same number of electrons, protons and neutrons as it is clear from the following explanation.

For N

2

No. of electrons in N

2

= 7 + 7 = 14

No. of protons in N

2

= 7 + 7 = 14

No. of neutrons = 7 + 7 = 14

For CO number of electrons

in C = 6

No. of electrons in O = 8

Total no. of protons = 6 + 8 = 14

No. of neutrons in C = 6

No. of neutrons in O = 8

Total no. of neutrons = 6 + 8 = 14

Q.13 How many molecule, of water are in 12 gram of ice?

Ans.

Mass of ice (water) = 12.0 gm

Molar mass of water = 18 g/mol

=

=

No. of molecules of water = 0.66 x 6.02 x 10

23

= 3.97 x 10

23

Q.14 Differentiate between limiting and non–limiting reactant ?

Ans. Limiting Reactant:

A limiting reactant is a reactant and that controls the amount of the product formed in a chemical reaction.

Non–Limiting Reactant:

The reactant which produces the excess amount of the product is called non–limiting reactant.

Q.15 Distinguish between actual yield and theoretical yield ?

Ans. Actual Yield:

The amount of the products obtained in a chemical reaction is called actual yield based on experiment.

Theoretical (Experiment) Yield:

The amount of the products calculated from the balanced chemical equation is called theoretical yield.

Q.16 What do you mean by percent yield? Give its significance ?

Ans.

The yield which is obtained by dividing actual yield with theoretical yield and multiplying by 100 is called percent yield.

% yield = x 100

Significance:

(i) % yield indicates the efficiency of reaction.

Q.17 Why actual yield is less than the theoretical yield?

Ans.

(a) Side reaction may takes place

(b) All the reactant may not be converted into products

(c) Mecahanical loss may occur like during

e.g Filtration, evaporation, crystallization, distillation etc.

Q.18 Calculate the mass of 10

–3

moles of MgSO

4

.

Ans.

MgSO

4

is an ionic compound. We will consider its formula mass instead of molecular mass.

Number of moles of substance

=

Formula mass of MgSO

4

= 120 gm/ml

Number of moles of MgSO

4

= 10

–3

moles Applying formula

10

–3

=

Mass of MgSO

4

= 120 x 10

–3

= 0.12 moles

Q.19 Define Avogadro’s number ?

Ans.

Avogadro’s number is the number of atoms, molecules and ions in one gram atom of an element, one gram molecule of a compound and one gram ion of substance, respectively. It is equal to 6.02 x 10

23

.

Ans.

The molecular mass of a substance expressed in grams is called molecule or gram mole or simply the mole of a substance. Moles of substance = 1 mole of water = 18.0 g 1 mole of H

2

SO

4

= 98.0 g Q.21 Define isotopes ? Ans.

Atoms of the same element which have different masses but same atomic numbers are called isotopes. For example carbon has three isotopes.

12

C

6

13

C

6

14

C

6

and expressed as C–12, C–13 and C–14. Similarly hydrogen has three isotopes H H H called protium, deuterium and tritium.

Q.22 Define (i) ions (ii) Positive ion (iii) Negative ion.

Ans. Ion

As specie having positive or negative charges are called ions. For example Cl

–1

, NO, Na

+

, Ca

++

. Positive Ion (Cation):

A specie has +ve charge is called positive ion and attracted towards Cathode . For example Na

+

, K

++

, Ca

++

.

Negative Ion (Anion)

A specie which has negative charge is called negative ion and attracted towards anode . For example F

–1

,Cl

–1

,Br

–1

andS

–2

P

–3

,C

–4

,SO, Cr

2

O, CO.

Q.23 Define and explain the molecular ion ?

When a molecule loses or gains an electron, molecular ion is formed. For example CH

4+

, CO

+

, N

2+

. Cationic molecular, ions are more abundant than anionic ions.

The molecular ions find applications of in calculation of molecular mass of a compound. The molecular ions also help in the determination of structure of macro molecules.

The break down of molecular ions obtained from the natural products can give important information about their structure.

Q.24 What do understand by the relative atomic mass ?

Ans.

Relative atomic mass is the mass of an atom of element as compared to the mass of an atom of carbon taken as 12.

The unit used to express the relative atomic mass is called atomic mass unit (amu). It is th of the mass of one carbon atom. The relative atomic mass of

12

C

6

is 12.00 amu. The relative atomic mass of H is 1.0078 amu.

Q.25 Define Gram atom ?

Ans.

The atomic mass of an element expressed in grams is called gram atom of an element.

Number of gram atoms of a meter an element

=

For example 1 gram atom of hydrogen = 1.008 gm

1 gram atom of carbon = 12.00 gm

1 gram at of uranium = 238 gm

Q.26 Define gram ion ?

Ans.

Number of gram ions =

1 gram ion of OH

–1

= 17 grams 1 gram ion of SO = 96 gram

1 gram ion of CO = 60 gram

Q.27 Define gram formula and moles ?

Ans.

The formula mass of an ionic compound expressed in grams is called gram formula of the substance.

Number of gram formula or moles of a substance

=

1 gram formula of NaCl = 58.50 gms

1 gram formula of Na

2

CO

3

= 106 gm

1 gram formula of AgNO

3

= 170 gm

The atomic mass, molecular mass, formula mass or ionic mass of the substance expressed in grams is called moles of those substances.

Q.28 Define molar volume ?

Ans.

The volume occupied by one mole of an ideal gas at standard temperature and pressure (STP) is called molar volume. The volume is equal to 22.414 dm

3

.

Q.29 Define and explain atomicity ?

Ans.

The number of atoms present in a molecule is called the atomicity. The molecule can be monoatomic, diatomic and triatomic etc. If the molecule contains one atom it is monoatomic, if it contains two atoms it is diatomic, and if it contains three atoms it is triatomic. Molecules of elements may contain one two or more

same type of atoms. For example He, Cl

2

, O

3

, P

4

, S

8

. The molecules of compounds consist of different kind of atoms. For example HCl, NH

3

, H

2

SO

4

, C

6

H

12

O

6

.

Q.30 Define an atom and molecule ?

Ans. Atom:

Atom is now defined as the smallest particle of an element, which may or may not have independent existence. For example He and Ne atoms have independent existence. While atoms of hydrogen, nitrogen and oxygen do not exist independently.

Molecule:

A molecule is the smallest particle of a pure substance( element or Compound ) which can exist independentally. For example N

2

, O

2

, Cl

2

, HCl, NH

3

and H

2

SO

4

are examples of molecules.

Q.31 What do you mean by empirical formula and molecular formula? How they are related to each

other ?

Ans. Empirical Formula:

It is the simplest formula that gives information about the simple ratio of atoms present in a compound.

In an empirical formula of a compound A

x

B

y

, there are X atoms of an element A and y atoms of an element B.

Molecular Formula:

The formula of a substance which is based on the actual molecule is called molecular formula. It gives the usual number of atoms present in the molecule. For example molecular formula of benzene is C

6

H

6

, while that of glucose is C

6

H

12

O

6

. The molecular formula and empirical formula are related to each other by the following relationship.

Molecular formula = n x (Empirical formula)

Where “n” is simple integer.

Ans.

There are many compounds, whose empirical formulas and molecular formulas are the same. For example H

2

O, CO

2

, NH

3

and C

12

H

22

O

11

have the same empirical and molecular formulas. Their simple multiple n is unity. Actually value of “n” is the ratio of molecular mass and empirical formula mass.

n =

Q.33 Ethylene glycol is used in automobile antifreeze. It has 38.7% carbon, 9.7% hydrogen and 51.6%

oxygen. Its molar mass is 62 gms mole

–1

. Determine its empirical and molecular formula ?

Ans.

C = 38.7%, H = 9.7%, O = 51.6%

Dividing above %ages by atomic mass.

We get molar ratios

C = = 3.225

H = = 9.7

O = = 3.225

Dividing above molar ratio by least ratio we get atomic ratio.

C = = 1

H = = 3

O = = 1

Empirical formula is CH

3

O

Molar mass = 62

Empirical formula mass = 12 + 3 + 16 = 31

Now

= = 2

Molar formula = n x Empirical formula

= 2 x CH

3

O

Molecular formula = C

2

H

6

O

2

Hence molecular formula of Ethylene glycol = C

2

H

6

O

2

Q.34 The combustion analysis of an organic compound shows it to contain 65.44% carbon 5.5%

hydrogen and 29.06% of oxygen. What is the empirical formula of the compound? If the molecular mass of the compound is 110.15. Calculate the molecular formula of the compound.

Ans.

First of all divide the percentage of each element by its atomic mass to get the number of from atoms or moles.

No. of gram atoms of carbon =

= 5.45 gram atoms of C

No. of gram atoms of hydrogen =

= 5.45 gram atoms of H

No. of gram atoms of oxygen =

= 1.82 gram atoms of 0

Mole ratio C : H : O

4.45 5.45 1.82

Divide number of grams atoms by the smallest number

C : H : O

: :

Carbon, hydrogen and oxygen are present in the given organic compound in ratio of 3 : 3 : 1. So the empirical formula is C

3

H

3

O.

In order to calculate the molecular formula first calculate the empirical formula mass.

Empirical formula mass = 3 x 12 + 3 x 1 + 16

= 36 + 3 + 16 = 55.05

Molar mass of the compound = 110.15

 = = = 2

Molecular formula = n x empirical formula

= 2 x C

3

H

3

O

= C

6

H

6

O

2

Q.35 Give relationships, between the amounts of substances and number of particles. There are three

useful relationships ?

Ans.

1. Number of atoms of an element = x N

A

2. Number of molecules of a compound

= x N

A

3. Number of ions of ionic species = x N

A

N

A

is the Avogadro’s number. The value is 6.02 x 10

23

.

Q.36 What are the types of relationships of stoichiometric calculations ?

Ans.

There are three types of relationships of stoichiometric calculations.

The relationship in which the mass of one substance is given and the mass of other substance is calculated.

2. Mass–mole or mole–mass relationship

The relationship in which mass of one substance is given and moles of other substance is to be calculated or vice versa.

3. Mass–volume or volume mass relationship

The relationship in which the mass of one substance is given and the volume of other substance is to be calculated or vice versa.

Q.37 Law of conservation of mass has to be obeyed during the stoichiometric calculations ?

Ans.

Stoichiometric calculations are based on balanced chemical equation and equation is balanced on the basis of Law of conservation of mass e.g

C+O2→ CO

In this equation stoichiometric calculations are not possible because it is not a balanced equation and it is not obeying Law of coseravtion.

TEXT BOOK EXERCISE

Q1. Select the most suitable answer from the given ones in each question. (i) The mass of one mole of electrons is

(b) Arrangement of electrons in orbital

(c) Chemical properties

(d) The extent to which they may be affected in electromagnetic field

(ii) Which of the following statements is not true?

(a) isotopes with even atomic masses are comparatively abundant

(b) isotopes with odd atomic masses and even atomic number are comparatively abundant

(c) atomic masses are average masses of isotopes.

(d) Atomic masses are average masses of isotopes proportional to their relative abundance

(iii) Many elements have fractional atomic masses, this is because

(a) The mass of the atom is itself fractional

(b) Atomic masses are average masses of isobars

(c) Atomic masses are average masses of isotopes.

(d) Atomic masses are average masses of isotopes proportional to their relative abundance

(iv) The mass of one mole of electrons is

(a) 008mg (b) 0.55mg (c) 0.184mg (d) 1.673mg

(v) 27g of Al will react completely with how much mass of O2 to produce Al2O3

(a) 8g of oxygen (b) 16g of oxygen

(c) 32g of oxygen (d) 24g of oxygen

(vi) The number of moles of CO2 which contain 8.0 g of oxygen.

(a) 0.25 (b) 0.50 (c) 1.0 (d) 1.50

(vii) The largest number of molecules are present in

(a) 3.6g of H2 O (b) 4.8g of C2H5 OH

(c) 2.8 g of CO (d) 5.4g of N2O5

(viii) One mole of SO2 contains

(a) 6.02x1023 atoms of oxygen

(b) 18.1x1023 Molecules of SO₂

(c) 6.02x1023 atoms of sulphur

(ix) The volume occupied by 1.4 g of N2at STP is

(a) 2.24 dm3 (b) 22.4dm3

(c) 1.12 dm3 (d) 112 cm3

(x) A limiting reactant is the one which

(a) is taken in lesser quantity in grams as compared to other reactants (b) is taken in lesser quantity in volume as compared to the other reactants (c) give the maximum amount of the product which is required

(d) give the minimum amount of the product under consideration

Ans: (i)a (ii)d (iii)d (iv)b (v)d (vi)a (vii)a (viii)c (ix)c (x)d

Q2: Fill in the blanks:

(i) The unit of relative atomic mass

is---(ii) The exact masses of isotopes can be determined by ---spectrograph.

(iii) The phenomenon of isotopes was first discovered by

---(iv) Empirical formula can be determined by combustion analysis for those compounds which

have---and ---in them.

(v) A limiting reagent is that which controls the quantities of

---(vi) I mole of glucose has---atoms of carbon ---of oxygen and ---of hydrogen.

(vii) 4g of CH₄ at OoC and I atm pressure has ---molecules of CH₄.

(viii) Stoichiometry calculations can by performed only when ---law is obeyed.

Ans: (i) amu (ii) mass (iii) Soddy (iv) carbon, hydrogen (v) Products (vi) 6NA,6NA,12NA

(vii) 1.505x1023 (viii) conservation and multiple proportion

Q3: Indicate true or false as the case may be:

(i) Neon has three isotopes and the fourth one with atomic mass 20.18 amu.

(ii) Empirical formula gives the information about he total number of atoms present in the molecule

(iii) During combustion analysis Mg(CIO₄)₂ is employed to absorb water vapors.

(iv) Molecular formula is the integral multiple of empirical formula and the integral multiple can never

be unity.

(vi) The number of electrons in the molecules of CO an dN₂ are 14 each, so 1 mg go each gas will

have same number of electrons.

(vii) Avogadro’s hypothesis is applicable to all types of gases, i.e., ideal and non-ideal .

(viii) Actual yield of a chemical reaction may by greater than the theoretical yield.

Ans. (i) False (ii) False (iii) True (iv) False

(v) False (vi) True (vii) False (viii) False

Q4: What are ions? Under What condition are they produced?

Ans: Ions can be produced by the following processes:

(i) By dissolving ionic compounds in water

(ii) By X-rays

(iii) In mass spectrometry

(iv) By removing or adding electron in atom

Q4:

(a) What are isotopes? How do you deduce the fractional atomic masses of elements form the relative isotopes abundance? Give two examples in support of your answer. ( See detail in Sublime subjective)

(b) How does a mass spectrograph show the relative aboundace of isotopes of an element? . ( See detail in Sublime subjective)

(c) What is the justification of two strong peaks in the mass spectrum for bromine; while for iodine only one peak at 127 amu , is indicated?

Ans The two strong peak in the mass spectrum for bromine represent two different isotopes of

bromine having nearly equal natural abundances. Only one peak at 127 amu in the mass spectrum for iodine indicates that it has only one isotope of atomic mass 127 amu.

Remember that!

Height of the peaks ∞ Relative abundance of isotopes

Q5: Silver has atomic number 47 and has 16 known isotopes but two occur naturally I,e, Ag -_____107 . and Ag _____109 . Given the following mass spectrometric data, calculated the average atomic mass of silver,

Isotopes mass (amu) percentage abundance

107

Ag 106.90509 51.84

109 _{Ag 108.90476 48.16}

Solution: The mass contribution for silver are:

Isotopes Fractional abundance isotopic mass mass contribution

107_{Ag 107 0.5184x107=55.4688}

109Ag 107 0.4816x109=52.4944

Fractional atomic mass of silver =107.9632

Hence the fractional atomic mass of silver is =107.9632 Ans.

Q6: Boron with atomic number 5 has two naturally occurring isotopes. Calculate the percentage

abundance of 10B and 11B from the following information.

Average atomic mass of boron =10.81 amu Isotopic mass of 10B =10.0129 amu Isotopic mass of 11B =11.0093

Solution: Let, the fractional abundance of 10B =x

The fractional abundance of 11B =1-x

Remember that the sum of the fractional abundances of isotopes must be equal to one, now,

The equation to determine the atomic mass of element is

(fractional abundance) (isotopic mass) (fractional abundance of 10B)(isotopic mass of10B

)+(fractional abundance of 11B) (isotopic mass of 11B)

=Average atomic mass of Boron (x)(10.0129)+(1-x)(11.0093) =10.81

10.0129x+11.00093x =10.81

10.0129x-11.00093x =10.81-11.0093 -0.9964x =-0.1993

x =

Fractional abundance of 10B =0.2000

Fractional abundance of 11B =(1-0.2000)=0.8000

By percentage the fractional abundance of isotope is

%of 10B =0.2000x100 =20% Answer

% of 11B =0.8000x100 =80%Answer

Q7: Define the following terms and give three examples of each.

(i) Gram atom (ii) Gram molecular mass

(iii) Gram molecular mass (iv) Gram ion (v)

molar volume (vi) Avogadro’s number (vii)

Stoichiometry (viii) Percentage yield

Q8: Justify the following statements:

(a) 23 g of sodium and 238g of uranium have equal number of atoms in the (b) Mg atom is twice heavier than that of carbon

(c) 180g of glucose and 342 g of sucrose have the same number of molecules but different number of atoms present in them.

(d) 4.9g of H2 SO4 when completely ionized in water , have equal number of positive and negative

charges but the number of positively charged ions are twice the number of negatively charged ions.

(e) One mg of K2 Cr O4 has thrice the number of ions than the number of formula units when ionized

in water.

(f) Two grams of H2 , 16 g of ch4 and 44g of CO2 occupy separately the volumes of 22.414 dm3 ,

although the sizes and masses of molecules of three gases are very different from each other.

Solution:

(a) 23g of Na =1 mole of Na =6.02x1023 atoms of Na

238g of U =1 mole of U =6.02x1023 atoms of U.

Since equal number of gram atoms(moles) of different elements contain equal number of atoms. Hence, 1 mole (23g ) of sodium and 1 mole (238)g of uranium contain equal number

(b) Since the atomic mass of Mg (24) is twice the atomic mass of carbon (12) therefore, Mg atom is twice heavier than that of carbon. Or

Mass of 1 atom of Mg= Mass of 1 atom of C =

Since the mass of one atom of Mg is twice the mass of one atom of C, therefore, Mg atom is twice heavier than that of carbon.

(c) 180 g of glucose = 1 mole of glucose =6.02x1023 molecules of glucose 342 g of

sucrose=1mole of sucrose =6.02x1023 molecules of sucrose

Since one mole of different compounds has the same number of molecules.

Therefore 1 mole (180g) of glucose and I mole (342g) of sucrose contain the same number

(6.02x1023)of molecules. Because one molecule of glucose, C6H12O6 contains 45 atoms

whereas one molecules of glucose, C12 H22 O11 contains 24 atoms. Therefore,

6.02x1023molecules of glucose contain different atoms as compound to6.02x1023 molecules of

sucrose. Hence, 180 g of glucose and 342g og sucrose have the same number of molecules but different number of atoms present in them.

(d) H2 SO4 2H+ + SO

When one molecules of H2 SO4 completely ionizes in water it produces two H+ ion and

one SO ion,. Hydrogen ion carries a unit positive charge whereas SO ion carries a double

negative charge. To keep the neutrality, the number of hydrogen are twice than the number of

soleplate ions. Similarly the ions produced by complete ionization of 4.8g of H2SO4 in water will

have equal number of positive and negative but the number of positively charged ions are twice the number of negatively charged ions.

(e) H2 SO4 2H+ + SO

K2 Cr O4 when ionizes in water produces two k+ ions one C O ion. Thus each formula

unit of K2 Cr O4produces three ions in solution .Hence one mg of K2 Cr O4 has thrice the number

of ion than the number of formula units ionized in water.

(f) 2g of H2 =1 mole of H2 =6.02x1023 molecules of H2 at STP =22.414dm3 16g of

CH4=1mole of CH4 =6.02x1023 molecules of CH4 at STP =22.414dm3 144g of CO2 =1mole of

Although H2 , CH4 and CO2 have different masses but they have the same number of

moles and molecules . Hence the same number of moles or the same number of molecules of

different gases occupy the same volume at STP. Hence 2 g of H2 ,16g of CH4 and 44 g of

CO2 occupy the same volume 22.414 dm3 at STP. The masses and the sizes of the molecules

do not affect the volumes.

Q10: Calculate each of the following quantities

(a) Mass in grams of 2.74 moles of KMnO₄ .

(b) Moles of O atoms in 9.0g of Mg (NO₃)₂ .

(c) Number of O atoms in 10.037g of Cu SO₄ .5H₂ O.

(d) Mass in kilograms of 2.6x 1020 molecules of SO₂ .

(e) Moles of C1 atoms in 0.822g C₂H₄C1₂ .

(f) Mass in grams of 5.136 moles of silver carbonate .

(g) Mass in grams of 2.78x1021 molecules of CrO₂ C1₂ .

(h) Number of moles and formula units in 100g of KC1O₃ .

(i) Number of K+ ions C1O ions, C1 atoms, and O atoms in (h)

Solution:

(a) No of moles of KMnO4 =2.74moles

formula mass of KMnO4 =39+55+64=158gmol -1

Mass of KMnO4 =?

Formula used:

Mass of KMnO4 = no of mole of KMnO4 x formula mass of KMnO4

=2.74 mol x 158 g mol-1

=432.92g Answer (b) Mass of Mg (NO3)2 =9g

Formula mass of Mg (NO3)2 =24+28+96=148g mol -1

No of moles of O atoms =? Formula used:

No of mole of Mg (NO3)2 =

0..06 moles of Mg (NO3)2contains =6x0.6

=0.36 moles of O atoms Alternatively,

148g of Mg (NO3)2 contains =6moles of O atoms

g of Mg (NO3)2contains =

=0.36 mole Answer

(c) Mass of CuSO4. 5H2O=10.037g

Formula mass of CuSO4_{. 5H}

2O=63.54+32+64+90

=249.546g mol -1

No of moles of CuSO4. 5H2O =?

No of moles of CuSO4. 5H2O =

=

Now, 1 mole of CuSO4 .5H2O contains 9moles of O atoms

0.04 mole of CuSO4 .5H2O contains=9x0.04

=0.36 moles of O atoms

Now, I mole of O atoms contains =6.02x1023 O atoms

0.36 mole of O atoms contains =6.02x1023 x0.36 oxygen atoms

=2.17x1023 oxygen atoms =2.17x1023 atoms Answer

(d) No of molecules of SO2 . =2.6x1020 molecules

Molecular mass of SO2 . =32+32=64 g mol-1

Now, Avogadro’s number , NA =6.02x1023 molecules of SO2

Mass of SO2 molecules =27.64x10-3 g = =27.64x10-6 kg =2.764x10-3 kg Answer

(e) Mass of C2 H4C1 = 0.822g

Molecular mass of C2 H4C1 =24+4+71=99 g mol-1

No of moles of C2 H4C1 =

Now, 1 mole of C2 H4C1 contains =2moles of C1 atoms

8.3x10-3mole of C2 H4C1 contains =2x8.3x10-3 mole of atom

=16.6x10-3

=0.0166mole of C1 atom =0.017 mole Answer

(f) No of mole of Ag2 CO3 =5.136moles

Formula mass of Ag2 CO3 =215.736+12+48=275.736 g mol-1

Mass of Ag2 CO3=No of moles of Ag2 CO3xformula mass of Ag2 CO3

=5.136molx275.736 g mol-1 =416.18g

=1416.2 g Answer

(g) Molecular mass of CrO2C12 =52+32+71=155g mol-1

_N

A =6.02x1023 molecules mol-1

Molecules of CrO2C12==2.78x1021 molecules

Now, mass of CrO2C12

= =

=71.578x10-2 g =0.71578

=0.716 g Answer (h) Mass of KCIO3 =100g

Formula mass of KCIO3 =39x35.5+48=122g mol-1

No of moles of KCIO3 =?

No of moles of KCIO3 =

No of formula units No of moles x Avogadro,s No

=0.816mole x 6.02x1023 formula units =4.91x1023 formula units

(i) No of K+ ions =4.91x1023 Answer

No of CIO ions =4.91x1023 Answer

No of CIO ions =4.91x1023 Answer

No of O atoms = 4.91x1023 x3

=14.73x1023 =1.473x1024 Answer

Q 11 Aspartame he artificial sweetener, has a molecular formula of C14 H18 N2O5 .

(a) What is the mass of one mole of aspartame? (b) How many moles are present in 52g of aspartame?

(c) What is the mass in grams of 10.122 moles of aspartame? (d) How many hydrogen atoms are present in 2.34g of aspartame?

(a) Molecular mass of aspartame =168+18+28+80=295g mol-1

Mass of 1 mole of aspartame =294g mol-1 Answer

(b) Mass of aspartame =52g

Molecular mass of aspartame =294g mol-1

No of moles of aspartame = =

=0.1768 mol

=0.177 mol Answer (c) No moles of aspartame = 10.122 moles

Molecular mass of aspartame =294g mol-1

Mass of aspartame =No of moles x Molar mass =10.122mol x 294g mol-1 =2975.87 g Answer (d) Mass of aspartame =243g

No of molecules of aspartame=?

No of molecules of aspartame= xNA

= =

=4.98x1021 molecules. Now,1 molecule of aspartame contains =18 H atoms

4.98x 1022 molecules =18x4.98x1021 H atoms

=89.64x1021H atoms

=8.964x1022 H atoms Answer

Q 12: A sample of 0.600 mole of a metal M reacts completely with excess of fluorine to from

46.8g MF2 .

(a) How many moles of F are present in the sample of MF2 that forms.

(b) which elements is represented by the symbol M ?

Solution:

(a) Formula of compound =MF2

No of moles of M =0.6 mol Mass of MF2 =46.8g

The molar of M:F in the compounds;

No of moles of F =0.6x2=1.2mol Answer

Mass of F =No of moles of Fx At . mass of F =1.2x19=22.8g Mass of compound =46.8g

Mass of metal, M =46.8-22.8 =24

= (b) The atomic mass of the elements, M =40

The metal is calcium, Ca Answer

Q 13 : In each pair , choose the larger of the indicated quantity ,or state if the samples are equal.

(a) Individual particles: 0.4 mole of oxygen molecules or0.4mole of oxygen atom. (b) Mass: 0.4 mole of ozone molecules or0.4mole of oxygen atoms

(c) Mass: 0.6 mole of C2 H4 or 0.6mole of 12

(d) Individual particles: 4.0g N2O4 or 3.3g SO2

(e) Total ions: 2.3 moles of NaC1O3 or 2.0mole of MgC12

(f) Molecules: 11.0g of H2Oor 11.0g H2O2

(g) Na+ ion: 0.500 moles of NaBr or 0.0145kg NaC1

(h) Mass: 6.02x1023 atoms of 235U or 6.02x1023 atoms of 238U

Ans:

(a) Number of molecules =moles x NA

Number of O2 molecules =0.4x6.02x1023 =2.408x1023 molecules

No of O atoms=0.4x6.02x1023=2.108x1023 atoms

There are equal number of individual particles in 0.4 mole of oxygen molecules and 0.4 mole of oxygen atom. In general, equal number of moles of different substances contains equal number of particles.

Both are equal Answer

(b) Mass of substance = moles x molar mass Mass of oxygen atoms =0.4x16=64g

Mass of ozone, O3 molecules =0.4x48=19.2g

0.4 moles of ozone molecules have larger mass than 0.4mole of oxygen atoms.

Ozone Answer

(c) Mass of C2H4 =0.6x28=1.68g

0.6mole of 12 have larger mass than 0.6 mole of C2H4

12 Answers

(d) No of molecules =

No of molecules in N2 O4 = x6.02x1023 =2.62 x1023 molecules

No of molecules in SO2 =x6.02x1023 =3.1x1022 molecules

3.3g of SO2 have larger number of individual particles than 4.0 g of N2 O4 .

SO2 Answer

(e) No of formula units =Moles x NA

No of formula units of NaC1O3 =2.3x6.02x1023=1.38x1024 formula units

No of ions in 1 formula units of NaC1O3=2

Total no of ions in MgC12 =2x1.38x1023=2.76x1024 ions

No of formula units of MgC12 =2.0x6.02x1023 x3=3.6x1024 ions

No .of ions in one formula unit of MgC12 =3

Total no of ions in MgC12 =1.20x1024 x3=3.6x1024 ions

2.0moles of MgC12 contain lager number of total ions than 2.3 moles of NaC1o

3-MgC1Answer

(f) No of molecules = NA

No of molecules in H2 O2 = x6.02x1023=3.68x1023 molecules

No of molecules in H2 O2 = x6.02x1023=1.95x1023 molecules

11.0g of H2 O2contains larger number of molecules than 11.0g of H2 O2

H2 O2Answer

(g) No of formula units =moles xNA

No of formula units NaBr =0.5x6.02x1023=3.01x1023 formula units

One formula units o NaBr contain Na+ ions =1

3.01 x1023 formula unit of NaBr contains Na +ions =3.01x1023 Na+ ions

No of formula units of NaC1 = x6.02x1023 =1.49x1023formula units

1.49x1023 formula units of NaC1 contains =1.49x1023 Na+ ions

0.500 moles of NaBr contains lager number of Na+ ions than 0.0145kg ofNaC1.

NaBr Answer

(h) Mass of atoms of an element =

Mass of 235Uatoms =x6.02x1023 =235g

Mass of 238U atoms =x6.02x1023=238g

238_{U Answer}

Q 13:

(a) Calculate the percentage of nitrogen in the four important fertilizer i.e., (i)NH3 (ii)NH2CONH2(Urea) (iii)(NH4)2SO4 (iv)NH4 NO3

(b) Calculate the percentage of nitrogen and phosphorus in each of the following: (i) NH4H2PO4 (ii) (NH4)) PO4 (iii) (NH4)4 PO4

Solution: (a) Mol-mass of NH3 =14+4=17g Mass of N =14g % of N =x100 =82.35% Answer (b) Mol-mass of NH2 CONH2 =28+4+12+16=60g Mass of N =28g %of N =x100 =46.35% Answer (c) Mol-mass of (NH2 )2 SO4 =28+8+32+64=132g Mass of N =28g % of N =x100 =21.21% Answer (d) Mol-mass of (NH2 )2 SO4 =28+4+48=80g Mass of N =28g %of N =x100 =35% Answer

(I) Mol-mass of (NH2 )2 SO4 =14+6+31+64=115g Mass of N =14g Mass of P =31g %of N =x100=12.17% Answer %of P ==26.96% Answer (II) Mol-mass of ((NH2 )2 SO4 =28+9+31+64=132g Mass of N =28g Mass of P = %of N = =21.21% Answer %of P = =23.48% Answer (III) Mol-mass of (NH2 )2 SO4 =42+12+31+64=149g Mass of N =42g Mass of P =31g %of N = %of P =

Q 14: Glucose C6 H12 O6 is the most important nutrient in the cell for generating chemical

potential energy. Calculate the mass% of each element in glucose and determine the number of C,H and O atoms in 10.5g go the sample.

Solution: Mol-mass of glucose C6 H12 O6 =72+12+96=180g Mass of C =72 Mass of H =12 Mass of O =96 % of C = =40% Answer % of H = =6.66% Answer % of O = =53.33% Answer Mass of C6 H12 O6 =10.5g Mol-mass of C6 H12 O6 =180g

Mol-mass of =180g mol-1

No of moles of C6 H12 O6 =

No of molecules of glucose =No of moles x NA

=0.058 molx 6.02x1023 molecules mol-1

=0.35x1023 molecules

=3.5x1022 molecules Now, 1 molecule of glucose contains =6C-atoms

3.4x1022 molecules of glucose contains =6x3.5x1022 C-atoms

=21x1022 =2.1x1023 C atoms Answer

1 molecules of glucose contains =12H-atoms

3.5x1022 molecules glucose contains =12x3.5x1022

_=4.2x1023 _{H- atoms Answer}

1 molecule of glucose contains =6 O –atoms

3.5 x 1022 molecules of glucose contains =6x3.5x1022

=2.1x1023 O-atoms Answer

Q 16: Ethylene glycol is used as automobile antifreeze .It has 38.7% carbon, 9.7% hydrogen

and 51.6% oxygen. Its molar mass is 62.1 grams mol-1 .Determine its molecular formula.

Solution:

% of C=38.37 g % of H =9.7g % of O=51.6g

At. Mass of C=12g mol-1 At. Mass of H=1.008g mol-1 At. Mass of O =16g mol-1

No of moles of C = No of moles of H = No of moles of O =

Atomic ratio is obtained by dividing the moles with 3.23, which is the smallest ratio. C :H :O

1 :3 :1 Empirical formula =CH3 O

Empirical formula mass =31

n=

Molecular formula =2x CH3 O

=C2 H6 O2 Answer

Q 16: Serotonin (Molecular mass= 176g mol-1 ) is a compound that conducts nerve impulses in brain and muscles. It contains 68.2 % C, 6.86% H, and 9.08% O. What is its molecular formula? Solution: No of moles of C = No of moles of H = No of moles of N = No of moles of O = C : H : N : O Atomic Ratio : : : 10 : 12 : 2 : 1 Empirical formula =C10 H12 N2 O

Empirical formula mass =120+12+28+16=176g mol-1

Molecular mass =176g mol-1

_n=

Q17: An unknown metal M reacts with S to from a compound with a formula M2S3 .If 3.12 gof

M reacts with exactly 2.88 g of sulphur, what are the names of metal M and the compound M2 S3 .

Solution:

Formula of compound = M2 S3

Mass of M =3.12g Mass of S =2.88g Atomic mass of S =32g mol-1 No of moles of S =

No of moles of S = The molar ratio of M: S in the compound is : No of moles of M = =0.06 mole Now, No of moles of M = At. Mass M =

The mass of M used in the formation of M2S3 is 3.12g. The product M2S3 therefore also contains

3.12g of M, because mass is conserved. The amount of M before and after reaction must be the same. Since we know both the number of moles of M and the mass of M , we can cal calculate the atomic mass of M as follows:

At. Mass of M = =52 Atomic number, Z =52

Q19: The octane present in gasoline burns according to the following equation.

2C8 H18 (i) + 2502(g) 16CO 2(g) + 18H2O (i)

(a) How many moles of O₂ are needed to react fully with 4 moles of actane?

(b) How many moles of CO₂ can be produced from one mole of actane?

(c) How many moles of water are produced by the combustion of 6 moles of octane?

(d) If this reaction is to be used to synthesize 8 moles of CO₂ how many grams of oxygen are

needed? How many grams of octane will be used?

Solution:

4 moles

(a) 2 moles 25 moles

2 moles of C8 H18 =25 moles of O2

4 moles of C8 H18 =

=50moles of O2 Answer

(b) 1 moles

2C8 H18 (i) + 2502(g) 16CO 2(g) + 18H2O (i)

2 moles

Now, 2 moles of C8 H18 =16 moles of CO2

1 mole of C8 H18 =

=8 moles of CO2 Answer

(c) 6 moles

2C8 H18 (i) + 2502(g) 16CO 2(g) + 18H2O (i)

2 moles

Now, 2 moles of C8 H18 =18 moles of H2 O(i)

6 moles of C8 H18 =

=54 moles of H2 O

(d) 6 moles

2C8 H18 (i) + 2502(g) 16CO 2(g) + 18H2O (i)

2 moles 1800moles

Now, 16 moles of CO2 =25 moles of O2

8 moles of CO2 =

=12.5 moles of CO2

Mol-mass of O2 =32g mol-1

=12.5 molx 32g mol-1

=400g of O2

Now, 16moles of CO2 =2moles of C8 H18

8 moles of CO2 =

=1 mole of C8 H18

Mass of C8 H18 =No of moles of C8 H18xMol.mass

ofC8H18 =1 molx 114 g mol-1

114g Answer

Q19: Calculate the number of grams of A12 S3 which can be prepared by the reaction of 20

g of A1 and 30 g of sulphur. How much the non-limiting reaction is in excess?

Solution:

Mass of A1 =20g

Molar mass of A1 =27g mol-1

No of moles of A1 = Mass of S = 30g

Molar mass of S =32g mol-1

_{No of moles of S =}

0.74 mole 0.94 mole

2A1 + 3S A12 S3

2 mole 3 mole 1 mole

Now, 2 moles of A1 =1 mole of A12 S3

0.74 moles of A1 =

=0.37 mole of A12 S3

Now, 3 moles of S =1 moles of A12 S3

0.94 moles of S =

=0.313 mole of A12 S3

Since S give the least number of moles of A12 S3 therefore, it is the limiting reactant.

No of moles of A12 S3 =0.313 mole

Molar mass of A12 S3 =150g mol-1

Mass of A12 S3=No of moles of A12 S3xMolar mass of A12 S3

=0.313molx 150 g mol-1 =46.95 g of A12 S3 Answer

The non-limiting reactant is A1 which is in excess. Now mass of A1 required reacting completely with 0.94 moles of S can be calculated as:

0.94 mole

2A1 + 3S A12 S3

2 mole 3 mole

Now, 3 moles of S =2 moles of A1 0.94 moles of S =

=

Mass of A1 =No of moles of A1 x molar mass of A1 =0.63x 27

=17g of A1 Mass of A1available =20g

Mass of A1 which reacts completely =17g with available S Excess of A1 =20-17=3g

Q20: A mixture of two liquids, hydrazine N2H4 and N2 O4 are used as a fuel in rockets. They

produce N2 and water vapors. How many grams of N2 gas will be formed by reacting 100g of

N2 O4 and 200g g of N2 O4.

2N2H4 + N2O2 3N2 +4 H2O

Solution:

Mass of2N2H4 =100g

Mass of N2O2 =200g

Molar mass of 2N2H4 =28+4=32g mol-1

Molar mass of N2O2 =28+64=92g mol-1

No of moles of N2H4 =

No of moles of N2O2 = 3.125moles 2.174 moles

2N2H4 + N2O2 3N2 +4 H2O

2 moles 1mole 3moles

Now , 2moles of N2H4 =3moles of N2 3.125moles of N2H4 =

Now , 1 mole of N2O2 =3moles of N2

2.174 moles of N2O4 =

=6.52 mole of N2O2

Since N2H4gives the least number of moles of N2, hence it is the limiting reactant.

Amount of N2 produced =4.69 moles

Molar mass of N2 =28g mol-1

_{Mass of N}

2 =4.69g molx 28g mol-1

=131032 g Answer

Q21: Silicon carbide (SiC) is an important ceramic material . It is produced by allowing sand

(SiO2 )to react with carbon at high temperature.

SiO2 + 3C SiC + 2CO

When 100kg sand isn reacted with excess of carbon, 51.4 kg of Sic is produced.

Solution:

Mass of SiO2 =100 kg=100000g

Mass of SiC produced =5.14 kg =51400g 100000g

SiO2 + 3C SiC + 2CO

60g 40g

Now, 60g of SiO2 =40g of SiC

100000g of SiO2 =

=66666.67 g Actual yield of Sic =51400 g

Theoretical yield of SiC =66666.67g % yield =

= =77.1%

Q22: (a) What is Stoichiometry? Give its assumptions? Mention two important law, which help to perform