CS– 1 STOICHIOMETRY - Einstein Classeseinsteinclasses.com/Stoichiometry.pdf · STOICHIOMETRY C1 In this chapter ... In aq. solution n = MV[M - molarity, V ... It is used in acid

CS– 1

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STOICHIOMETRY

C1 In this chapter we will discuss the calculations based on chemical equations. It has been classified into two

parts :

1. Mole Concept

2. Equivalent Concept

C2 MOLE CONCEPT :

In mole concept we deal with different types of relations like weight-weight, weight-volume, orvolume-volume relationship between reactants or products of the reaction.

Mole concept is based on balanced chemical chemical reaction. Some basic definitions used in moleconcept are as follows :

Limiting Reagent : A reagent which is consumed completely during the chemical reaction.

weightmolecularoratomic

substanceofweightn)substance(aofmolesofNumber

Also, number Avogadro

molecules ofnumber Givenn)substance(aofmolesofNumber

In gas phase reaction number of moles of a gas (n) = RT

PV,

At STP/NTP one mole of any gas contains 22.4 L i.e. at 273 K and 1 atm pressure.

In aq. solution n = MV [M - molarity, V - volume of solution]

Practice Problems :

1. Chlorine can be produced by reacting H2SO

4 acid with a mixture of MnO

2 and NaCl. The reactions

follows the equation : 2NaCl + MnO2 + 3H

2SO

4 2NaHSO

4 + MnSO

4 + Cl

2 + H

2O what volume of

chlorine at STP can be produced from 100 g of NaCl ? (At. wt. Na = 23, Cl = 35.5)

(a) 19.15 lt (b) 30 lt (c) 29 lt (d) 5 lt

2. A solution contains 5 g of KOH was poured into a solution containing 6.8 g of AlCl3, find the mass of

precipitate formed [At. wt. : H-1, Al-27, Cl-35.5, K-39]

(a) 2.3 g (b) 23 g (c) 32 g (d) 0.32 g

[Answers : (1) a (2) a]

C3 DIFFERENT WAYS OF EXPRESSING THE CONCENTRATION TERMS :

Important Definitions :

100solutionofmass

soluteofmasspercentmass

L in solution of Vol.

solute of moles of No. )Molarity(M , unit of molarity are mol/lit., M or molar..

L in solution of Vol.

solute of lentsgramequiva of No. )N(Normality , unit of normality are g-eq./lit., N or normal.

solventof.wt

soluteofmolesof.No)m(Molality , unit of molality are mol/kg, m or molal.

BA

AA

nn

n)(fractionMole

x .

610solutionofmass

soluteofmassppm

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weight Equivalent

solute ofWeight )solute(n of sequivalent gram of .No eq

factorn

weightIonic)or(weightAtomic)or(weightMolecularweightEquivalent

The relation between different concentration terms :

1. neq

= nmol

× n-factor 2. neq

= Normality × Volume (L)

3. Number of moles(nmol

) = Molarity × Volume (L) 4. Normality = Molarity × n-factor

5.M

xd10M

6.

MMd1000

1000Mm

7.AB

B

M)x1(

1000xm

(d density of solution in g/ml, M molar mass of solute, xB and x

A are mole fraction of solute and

solvent respectively, MA molar mass of solvent)

Calculation of ‘n’ Factor for Different Compounds :

1. Acids : n = basicity

H3PO

4n = 3 H

3PO

3n = 2

H3PO

2n = 1 H

3BO

3n = 1

2. Bases : n = acidity of base

e.g. Ammonia and all amines are monoacidic bases,

NaOH(n = 1), Na2CO

3(aq) n = 2, NaHCO

3(n = 1)

3. Salt : (Which does not undergo redox reactions)

n factor = Total cationic or anionic charge, e.g. Na3PO

4 n = 3, Ba

3(PO

4)

2 n = 6

4. Oxidizing Agents or Reducing Agents : ‘n’ factor = change in oxidation number Ornumber of electron lost or gained from one mole of the compound.

C4 EQUIVALENT CONCEPT

It is based on law of equivalence which is explained as follows :

Law of chemical equivalents : In a chemical reaction the equivalents of all the species (reactants orproducts) are equal to each other provided none of these compounds is in excess.

N1V

1 = N

2V

2 (when normalities and volumes are given).

If the number of equivalence of both the reactants are different then reactant with the lesser number ofequivalence will be the limiting reagent.

Application of equivalent concept : It is used in acid base titration, back titration and double titration,similarly in redox titration. Equivalent concepts can be used on all reaction whether they are balanced ornot balanced but mole concept is used in solving the problems when the reactions are balanced.

Basic principles of tirations :

In voltmetric analysis, a given amount (weight or volume) of an unknown substance is allowed to react witha known volume of a standard solution slowly. A chemical reaction takes place between the solute of anunknown substance and the solute of the standard solution. The completion of the reaction is indicated bythe end point of the reaction, which is observed by the colour change either due to the indicator or due to thesolute itself. Whether the reactions during the analysis are either between an acid and or base or betweenO.A. and R.A., the law of equivalence is used at end point.

Following are the different important points regarding this process :

(i) In case of acid base titration at the equivalence point

(neq

)acid = (neq

)base

(ii) In case of redox titration

(neq

)oxidant = (neq

)reductant

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(iii) If a given volume of solution is diluted then number of moles or number of equivalence ofsolute remains same but molarity or normality of the solution decreases.

(iv) If a mixture contains more than one acids and is allowed to react completely with the base thenat the equivalence point, (n

eq) acid

1 + (neq) acid

2 + ... = (n

eq) base

(v) Similarly if a mixture contains more than one oxidising agents then at equivalence point,

(neq

) O.A1 + (n

eq) O.A

2 +... = (n

eq) reducing agent.

(vi) If it is a difficute to solve the problem through equivalence concept then use the mole concept.

Back titration :

This is a method in which a substance is taken in excess and some part of its has to react with anothersubstance and the remaining part has to be titrated against standard reagent.

Double titration :

This is a titration of specific compound using different indicators. Let us consider a solid mixture of NaOH,Na

2CO

3 and inert impurities.

When the solution containing NaOH and Na2CO

3 is titrated using phenolphalein indicator following reac-

tion takes place at the phenolphthalein end point –

NaOH + HCl NaCl + H2O

Na2CO

3 + HCl NaHCO

3 + H

2O

Here, eq. of HClof.eqCONaof.eq2

1NaOH

)2n(32

When methyl orange is used, Na2CO

3 is converted into NaCl + CO

2 + H

2O

Hence, eq. of NaOH + eq. of )2n(

32CONa

= eq. of HCl

TITRATION OF MIXTURE OF BASES WITH TWO INDICATORS

Every indicator has a working range

Indicator pH range Behaving as

Phenolphthalein 8 — 10 weak organic acid

Methyl orange 3 — 4.4 weak organic base

Thus methyl orange with lower pH range can indicate complete neutralisation of all typesof bases. Extent of reaction of different bases with acid (HCl) using these two indicatorssummarised below

Phenolphthalein Methyl Orange

NaOH 100% reaction is indicated 100 % reaction is indicated

NaOH + HCl NaCl + H2O NaOH + HCl NaCl + H

2O

Na2CO

350% reaction upto NaHCO

3100% reaction is indicated

stage is indicated Na2CO

3 + 2HCl 2NaCl + H

2O

Na2CO

3 + HCl NaHCO

3 + NaCl + CO

2

NaHCO3

No reaction is indicated NaHCO3 + HCl NaCl + H

2O +

CO2

100% reaction is indicated

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Species Changed to Reactions Electron exchangedor change in O.N.

1.

Eq. wt.

MnO4— (O.A.) Mn2+ in acidic

mediumMnO

4— + 8H+ + 5e— Mn2+ +

4H2O

5

5

ME

2. MnO4— (O.A.) MnO

2 in basic

mediumMnO

4— + 3e— + 2H

2O MnO

2 +

4OH—

33

ME

3. MnO4— (O.A.) MnO

42— in

neutral mediumMnO

4— + e— + 2H

2O MnO

22— 1

1

ME

4. Cr2O

72—(O.A.) Cr3+ in acidic

mediumCr

2O

72— + 14H+ + 6e— 2Cr3+ +

7H2O

66

ME

5. MnO2(O.A.) Mn2+ in acidic

mediumMnO

2 + 4H+ + 2e— Mn2+ +

2H2O

22

ME

6. Cl2(O.A.)

(in bleachingpowder)

Cl— Cl2 + 2e— 2Cl— 2

2

ME

7. CuSO4 (O.A.)

(in iodometrictitration)

Cu+ Cu2+ + e– Cu+ 11

ME

8. S2O

32— (R.A.) S

4O

62— 2S

2O

32— S

4O

62— + 2e— 2 (for two molecules) M

5

ME

9. H2O

2(O.A.) H

2O H

2O

2 + 2H+ + 2e— 2H

2O 2

2

ME

10. H2O

2(R.A.) O

2H

2O

2 O

2 + 2H+ + 2e—

(O.N. of oxygen in H2O

2 is (–1)

per atom)

22

ME

11. Fe2+ (R.A.) Fe3+ Fe2+ Fe3+ + e— 11

ME

Estimation of Reaction Relation between O.A. and R.A.

1. I2

I2 + 2Na

2S

2O

3 2NaI + Na

2S

4O

62—

I2 + 2S

2O

32— 2I— + S

4O

62—

I2 2I— 2Na

2S

2O

3

Eq. wt. (Na2S

2O

3) =

1

ME

2. CuSO4

2CuSO4 + 4KI Cu

2I

2 + 2K

2SO

4 + I

2

or 2Cu2+ + 4I— Cu2I

2 + I

2

white ppt.

2CuSO4 I

2 2Na

2S

2O

3

Eq. wt. of CuSO4 =

1

M

3. CaOCl2

CaOCl2 + H

2O Ca(OH)

2 + Cl

2

Cl2 + 2KI 2KCl + I

2

Cl2 + 2I— 2Cl— + I

2

CaOCl2 Cl

2 I

2 2I— 2Na

2S

2O

3

Eq. wt. of CaOCl2 =

2

M

4. MnO2 MnO

2 + 4HCl (conc.) MnCl

2 +

Cl2 + 2H

2O

Cl2 + 2KI 2KCl + I

2

or MnO2 + 4H+ + 2Cl— Mn2+ + 2H

2O

+ Cl2

Cl2 + 2I— I

2 + 2Cl—

MnO2 Cl

2 I

2 2I— 2Na

2S

2O

3

Eq. wt. of MnO2 =

2

M

5. IO3

— IO3— + 5I— + 6H+ 3I

2 + 3H

2O IO

3— 3I

2 6I 6Na

2S

2O

3

Eq. wt. IO3— =

6

M

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Practice Problems :

1. [Na+] in a solution prepared by mixing 30.00 mL of 0.12 M NaCl with 70 mL of 0.15 M Na2SO

4 is

(a) 0.135 M (b) 0.141 M (c) 0.210 M (d) 0.246 M

2. The equivalent mass of MnSO4 is half of its molar mass when it is converted to

(a) Mn2O

3(b) MnO

2(c) MnO

4– (d) MnO

42–

3. The anion nitrate can be converted into ammonium ion. The equivalent mass of NO3– ion in this

reaction would be

(a) 6.20 g (b) 7.75 g (c) 10.5 g (d) 21.0 g

4. When BrO3– ion reacts with Br– ion in acid solution Br

2 is liberated. The equivalent weight of KBrO

3

in this reaction is

(a) M/8 (b) M/3 (c) M/5 (d) M/6

5. The number of moles of KMnO4 that will be needed to react completely with one mole of ferrous

oxalate in acidic solution is

(a) 3/5 (b) 2/5 (c) 4/5 (d) 1

6. 5 ml of N-HCl, 20 ml of N/2-H2SO

4 and 30 ml of N/3 – HNO

3 are mixed together and the volume

made to 1 litre.

(i) The normality of the resulting solution is

(a) N/5 (b) N/10 (c) N/20 (d) N/40

(d)

(ii) The wt. of pure NaOH required to neutralize the above solution is

(a) 10 g (b) 2 g (c) 1 g (d) 2.5 g

7. 0.7 g of a sample of Na2CO

3.xH

2O were dissolved in water and the volume was made to 100 ml, 20 ml

of this solution required 19.8 ml of N/10 HCl for complete neutralization. The value of x is

(a) 7 (b) 3 (c) 2 (d) 5

6. H2O

2H

2O

2 + 2I— + 2H+ I

2 + 2H

2O H

2O

2 I

2 2I— 2Na

2S

2O

3

Eq. wt. H2O

2 =

2

M

7. Cl2

Cl2 + 2I— 2Cl— + I

2Cl

2 I

2 2I— 2Na

2S

2O

3

Eq. wt. of Cl2 =

2

M

8. O3

O3 + 6I— + 6H+ 3I

2 + 3H

2O + O

2O

3 3I

2

Eq. wt. of O3 =

2

M

9. ClO— ClO— + 2I— + 2H+ H2O + Cl— + I

2ClO— I

2 2I 2Na

2S

2O

3

Eq. wt. of ClO— = 2

M

10. Cr2O

72— Cr

2O

72— + 14H+ + 6I— 3I

2 + 2Cr3+ +

7H2O

Cr2O

72— 3I

2 6I—

Eq. wt. of Cr2O

72— =

6

M

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8. 100 mL of 1 M KMnO4 oxidised 100 mL of H

2O

2 in acidic medium (when MnO

4– is reduced to Mn2+);

volume of same KMnO4 required to oxidise 100 mL of H

2O

2 in basic medium (when MnO

4–. is

reduced to MnO2) will be

(a) mL3

100(b) mL

3

500(c) mL

3

300(d) 100 mL

9. 100 mL of a mixture of NaOH and Na2SO

4 is neutralised by 100 mL of 0.5 M H

2SO

4. Hence amount

of NaOH in 100 mL mixture is

(a) 0.2 g (b) 0.4 g (c) 0.6 g (d) 1.0 g

10. 3 mol of a mixture of FeSO4 and Fe

2(SO

4)

3 required 100 mL of 2 M KMnO

4 solution is acidic

medium. Hence mol fraction of FeSO4 in the mixture is

(a)3

1(b)

3

2(c)

5

2(d)

5

3

11. 5.3 g of M2CO

3 is dissolved in 150 mL of 1 N HCl. Unused acid required 100 mL of 0.5 N NaOH.

Hence equivalent weight of M is

(a) 23 (b) 12 (c) 24 (d) 13

[Answers : (1) d (2) b (3) b (4) c (5) a (6) (i) d (ii) c (7) c (8) b (9) b (10) a (11) a]

C5 VOLUME STRENGTH OF H2O

2

x volume of H2O

2 means x litre of O

2 is liberated by 1 volume of H

2O

2 on decomposition

STPatlit4.2222

gm6822 OOH2OH2

Volume strength of H2O

2 solution = N × 5.6...... (where N is the normality of the H

2O

2

solution

Practice Problems :

1. (a) Calculate the strength of ‘20 V’ of H2O

2 in terms of :

(i) normality (ii) grams per litre (iii) molarity and (iv) percentage

(b) Calculate the volume strength of 2.0 N H2O

2 solution.

2. In a 50 ml solution of H2O

2 an excess of KI and dilute H

2SO

4 were added. The I

2 so liberated required

20 ml of 0.1 N Na2S

2O

3 for complete reaction. Calculate the strength of H

2O

2 in grams per litre.

[Answers : (a) (i) 3.58 N (ii) 60.86 g/lit. (iii) 1.79 M (iv) 6.086% (W/V) (b) 11.2 V (2) 0.68 g/litre]

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SINGLE CORRECT CHOICE TYPE

1. 1 g of the carbonate of a metal was dissolved in25 ml of N-HCl. The resulting liquid required 5 mlof N-NaOH for neutralization. The eq. wt. of themetal carbonate is

(a) 50 (b) 30

(c) 20 (d) None

2. If 0.5 mol of BaCl2 is mixed with 0.20 mol of Na

3PO

4,

the maximum amount of Ba3(PO

4)

2 that can be

formed is

(a) 0.70 mol (b) 0.50 mol

(c) 0.20 mol (d) 0.10 mol

3. When one gram mol of KMnO4 reacts with HCl,

the volume of chlorine liberated at NTP will be

(a) 11.2 litres (b) 22.4 litres

(c) 44.8 litres (d) 56.0 litres

4. 34 g of hydrogen peroxide is present in 1120 ml ofsolution. This solution is called

(a) 10 vol solution (b) 20 vol solution

(c) 30 vol solution (d) 32 vol solution

5. To prepare a solution that is 0.50 M KCl startingwith 100 mL of 0.40 M KCl

(a) add 0.75 g KCl

(b) add 20 mL of water

(c) add 0.10 mol KCl

(d) evaporate 10 mL water

6. In hot alkaline solution, Br2 disproportionates to

Br– and BrO3–

3Br2 + 6OH– 5Br– + BrO

3– + 3H

2O

hence equivalent weight of Br2 is (molecular

weight = M)

(a)6

M(b)

5

M

(c)5

M3(d)

5

M5

7. When 80 mL of 0.20 M HCl is mixed with 120 mLof 0.15 M KOH, the resultant solution is the sameas a solution of

(a) 0.16 M KCl and 0.02 M HCl

(b) 0.08 M KCl

(c) 0.08 M KCl and 0.01 M KOH

(d) 0.08 M KCl and 0.01 M HCl

8. Molality of 18 M H2SO

4 (d = 1.8 gmL–1) is

(a) 36 mol kg–1 (b) 200 mol kg–1

(c) 500 mol kg–1 (d) 18 mol kg–1

9. 1 g equiv. of a substance is the weight of that amountof a substance which is equivalent to

(a) 0.25 mol of O2

(b) 0.50 mol of O2

(c) 1 mol of O2

(d) 8 mol of O2

10. The molality of a H2SO

4 solution is 9. The weight of

the solute in 1 kg H2SO

4 solution is

(a) 900.0 g (b) 468.65 g

(c) 882.0 g (d) 9.0 g

11. The density of 1 M solution of NaCl is 1.0585 g/mL.The molality of the solution is

(a) 1.0585 (b) 1.00

(c) 0.10 (d) 0.0585

12. Which is false about H3PO

2

(a) it is tribasic acid

(b) one mole is neutralised by 0.5 molCa(OH)

2

(c) NaH2PO

2 is normal salt

(d) it disproportionates to H3PO

3 and PH

3 on

heating.

13. When KMnO4 acts as an oxidising agent and

ultimately forms [MnO4], MnO

2, Mn

2O

3, Mn2+ then

the number of electrons transferred in each caserespectively is

(a) 3, 5, 7, 1 (b) 1, 5, 3, 7

(c) 4, 3, 1, 5 (d) 1, 3, 4, 5

14. With increase of temperature, which of thesechanges ?

(a) mole fraction

(b) Fraction of solute present in water

(c) molality

(d) weight fraction of solute

15. In a compound C, H and N atoms are present in9:1:3. 5 by weight of compound is 108. Molecularformula of compound is

(a) C2H

6N

2(b) C

3H

4N

(c) C9H

12N

3(d) C

6H

8N

2

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ANSWERS (SINGLE CORRECTCHOICE TYPE)

11. b

12. a

13. d

14. b

15. d

16. d

17. a

18. b

19. b

20. c

1. a

2. d

3. d

4. a

5. a

6. c

7. c

8. c

9. a

10. b

21. c

22. b

23. c

24. b

25. c

16. 8 g of sulphur are burnt to form SO2 which is

oxidised by Cl2 water. The solution is treated with

BaCl2 solution. The amount of BaSO

4 precipitated

is

(a) 1 mol (b) 0.5 mol

(c) 0.24 mol (d) 0.25 mol

17. One mole of a mixture of CO and CO2 requires

exactly 20 gram of NaOH in solution for completeconversion of all the CO

2 into Na

2CO

3. How many

grams of NaOH would it require for conversion intoNa

2CO

3 if the mixture (one mole) is completely

oxidised to CO2

(a) 60 grams (b) 80 grams

(c) 40 grams (d) 20 grams

18. One gram of a mixture of Na2CO

3 and NaHCO

3

consumes y gram equivalents of HCl for completeneutralisation. One gram of the mixture is stronglyheated, the cooled and the residue treated with HCl.How many gram equivalents of HCl would berequired for complete neutralization ?

(a) 2 y gram equivalent

(b) y gram equivalents

(c) 3y/4 gram equivalents

(d) 3y/2 gram equivalents

19. If equal volumes of 1 M KMnO4 and 1 M K

2Cr

2O

7

solutions are allowed to oxidise Fe (II) to Fe(III),then Fe (II) oxidised will be

(a) more by KMnO4

(b) more by K2Cr

2O

7

(c) equal in both cases

(d) none of these

20. 0.5 g of fuming H2S

2O

7(Oleum) is diluted with

water. This solution is completely neutralized by26.7 ml of 0.4 N NaOH. The percentage of free SO

3

in sample is

(a) 30.6% (b) 40.6%

(c) 20.6% (d) 50%

21. One mole of N2H

4 loses 10 mol of electrons to form

a new compound Y. Assuming that all the nitrogenappears in the new compound. What is theoxidation state of nitrogen in Y.

(a) – 1 (b) – 3

(c) + 3 (d) + 5

22. 25 ml of a solution of barium hydroxide ontitration with a 0.1 molar solution of hydrochloricacid gave a titre value of 35 ml. The molarity ofbarium hydroxide solution was

(a) 0.35 (b) 0.07

(c) 0.14 (d) 0.28

23. To neutralise completely 20 mL of 0.1 M aqueoussolution of phosphorus acid (H

3PO

3), the volume

of 0.1 M aqueous KOH solution required is

(a) 10 mL (b) 20 mL

(c) 40 mL (d) 60 mL

24. Excess of KI reacts with CuSO4 solution and then

Na2S

2O

3 solution is added to it. Which of the

statements is incorrect for this reaction ?

(a) Cu2I

2 formed

(b) CuI2 is formed

(c) Na2S

2O

3 is oxidised

(d) evolved I2 is reduced

25. Two solutions of a substance (non electrolyte) aremixed in the following manner 480 ml of 1.5 M firstsolution and 520 mL of 1.2 M second solution. Whatis the molarity of the final mixture ?

(a) 1.20 M (b) 1.50 M

(c) 1.344 M (d) 2.70 M

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EXCERCISE BASED ON NEW PATTERN

COMPREHENSION TYPEComprehension-1A 2.0 g sample of a mixture containing sodiumcarbonate, sodium bicarbonate and sodiumsulphate is gently heated till the evolution of CO

2

ceases. The volume of CO2 at 750 mm Hg pressure

and at 298 K is measured to be 123.9 ml. A 1.5 g ofthe same sample requires 150 ml of M/10 HCl forcomplete neutralisation.

1. The percentage composition of sodium carbonatein the mixture is(a) 42% (b) 18%(c) 53% (d) none

2. The percentage composition of sodiumbicarbonate in the mixture is(a) 42% (b) 18%(c) 53% (d) none

3. The amount of HCl used for completeneutralization of 2g of sample is(a) 0.73 (b) 0.42(c) 0.53 (d) 0.18Comprehension-2Reducing sugars are sometimes characterized by anumber R

Cu, which is defined as the number of mg

of copper reduced by 1 g of the sugar, in which thehalf-reaction for the copper isCu2+ + OH— Cu

2O + H

2O (unbalanced)

It is sometimes more convenient to determine thereducing power of a carbohydrate by an indirectmethod. In this method 43.2 mg of thecarbohydrate was oxidized by an excess ofK

3Fe(CN)

6. The Fe(CN)

64— formed in this reaction

required 5.29 cm3 of 0.0345 N Ce (SO4)

2 for

reoxidation to Fe(CN)6

3— [the normality of thecerium(IV) sulfate solution is given with respect tothe reduction of Ce4+ to Ce3+].

4. The number of milimole of Ce(SO4)

2 used for

re-oxidation to Fe(CN)63– is

(a) 0.183 (b) 0.0915(c) 1.83 (d) 9.14

5. 43.2 mg of sugar was reduced by(a) 11.01 mg Cu2+ (b) 11.6 mg Cu2+

(c) 1.101 mg Cu2+ (d) 1.16 mg Cu2+

6. The RCu

value for the sample is(a) 269 (b) 2.69(c) 26.9 (d) 0.269Comprehension-3An acid solution of a KReO

4 sample containing

26.83 mg of combined rhenium was reduced bypassage through a column of granulated zinc. Theeffluent, including was washing from the column,was then titrated with 0.1000 N KMnO

4; 11.45 mL

of the standard permanganate was required for thereoxidation of all the rhenium to the perrhenate ion,

ReO4

—. Assuming that rhenium was the onlyelement reduced.

7. Number of equivalence of KMnO4 used

(a) 1.145 (b) 1.145 × 10–3

(c) 11.45 × 10–3 (d) 3.25 × 10–3

8. What is the oxidation state to which rhenium wasreduced by the zinc column ?(a) 0 (b) –1(c) –2 (d) –3

MATRIX-MATCH TYPEMatching-1Column - A Column - B

(A) 0.1M of MnO4– (P) oxidised 0.25M

in acidic medium C2O

42–

(B) 0.6 mol of KMnO4

(Q) oxidised 0.5Min acidic medium Fe2+

(C) Molarity of pure water (R) oxidised(density of water = 1g/ml) 0.166M

FeC2O

4

(D) 0.083 molar Cr2O

72– (S) oxidises 1 mol

in acidic medium of Fe(C2O

4)

2

(T) 5.55 × 10Matching-2Column-A Column-B

(A) Total no. of electrons in (P) 3.01 × 1021

1.6 g methane are(B) No. of sulphate ions (Q) 6.02 × 1023

present in 50mL of 0.1MH

2SO

4 solution are

(C) 500cm3 of 0.2 molar NaCl (R) 6.02 × 1022

is added to 100cm3 of0.5 molar AgNO

3. Thus no.

of ions of AgCl formed are(D) The no. of of Fe2+ ions (S) 3.01 × 1022

formed when excess ofiron is treated with 5 mlof 0.04 N HCl

(T) 6.02 × 1020

MULTIPLE CORRECT CHOICE TYPE1. Assuming complete dissociation of H

2SO

4 as

(H2SO

4 2H+ + SO

42–), the number of sulphate ions

present in 50 ml of 0.1 M H2SO

4 solution are

(a) 5 × 10–3 mol (b) 3.01 × 1021

(c) 5 × 1023 (d) 3.01 × 10–3 mol2. In the following redox reaction 2MnO

4– + 10Cl– +

16H+ 2Mn2+ 5Cl2 + 8H

2O. Pick up the correct

statements.(a) MnO

4– is reduced

(b) Cl– is oxidising agent(c) MnO

4– is an oxidising agent

(d) Cl– is reduced

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3. Which of the following represents redox reaction(s) ?(a) Cu + Cu2+ 2Cu+

(b) MnO4– + Mn2+ + OH– MnO

2 + H

2O

(c) Cr2O

72– + 2OH– 2CrO

42– + H

2O

(d) 2CrO42– + 2H+ Cr

2O

72– + H

2O

4. Which of the following are redox reactions ?(a) Zn + 2HCl ZnCl

2 + H

2

(b) Al(OH)3 + 3HCl AlCl

3 + 3H

2O

(c) Disproportionation of Cu+ ions in a givensolution

(d) Ag+(aq.) + I–(aq.) Ag I (s)5. Which of the following is disproportionation ?

(a) 2Cu+ Cu + Cu2+

(b) 3Cl2 + 6OH– ClO

3– + 5Cl– + 3H

2O

(c) 2H2S + 8O

2 2H

2O + 3S

(d) Na + 2

1Cl

2 NaCl

6. Oxidation number of C is zero in(a) CHCl

3(b) CH

2Cl

2

(c) C6H

12O

6(d) CO

7. Among the species given below which can act asoxidising as well as reducing agent ?(a) SO

2(b) SO

3

(c) H2O

2(d) H

2S

8. Which of the following changes involve oxidation ?(a) change of Zn to ZnSO

4 by reaction with

H2SO

4

(b) change of Cl2 to chloride ion

(c) change of H2S to S

(d) change of sodium sulphite to sodiumsulphate.

Assertion-Reason TypeEach question contains STATEMENT-1 (Assertion)and STATEMENT-2 (Reason). Each question has4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.

(A) Statement-1 is True, Statement-2 is True;Statement-2 is a correct explanationfor Statement-1

(B) Statement-1 is True, Statement-2 is True;Statement-2 is NOT a correctexplanation for Statement-1

(C) Statement-1 is True, Statement-2 is False(D) Statement-1 is False, Statement-2 is True

1. STATEMENT-1 : 22.4 L of ethane at N.T.P.contains one mole of hydrogen molecules.STATEMENT-2 : One mole of hydrogenmolecules at N.T.P. occupies 22.4 L of volume.

2. STATEMENT-1 : The masses of oxygen whichcombine with fixed mass of nitrogen in N

2O, NO,

N2O

3 bears a simple multiple ratio.

STATEMENT-2 : The combination according tolaw of multiple proportions.

3. STATEMENT-1 : 8.075 × 10–2 kg of Glauber’s saltis dissolved in water to obtain 1 dm3 of solution ofdensity 1077.2 kg m–3. The molarity of theresultant solution is 0.25 M .STATEMENT-2 : The volume in mL of 0.5 MH

2SO

4 needed to dissolve 0.5 g of copper (II)

carbonate is 24.3 mL4. STATEMENT-1 : Basicity of an acid can change

in different reactions.STATEMENT-2 : As the equivalent weight alwaysremains same.

5. STATEMENT-1 : If equal volume of C M KMnO4

and C M K2Cr

2O

7 solutions are allowed to oxidised

Fe2+ to Fe3+ in acidic medium, then Fe2+ oxidisedby KMnO

4 is more then K

2Cr

2O

7 of same

concentrations.STATEMENT-2 : Number of moles of electronsgained by 1 mole of K

2Cr

2O

7 is more than the 1

mole of KMnO4

(Answers) EXCERCISE BASED ON NEW PATTERN

COMPREHENSION TYPE

1. c 2. a 3. a 4. b 5. b 6. a

7. b 8. b

MATRIX-MATCH TYPE

1. [A-P, Q, R ; B-S ; C-T ; D-P] 2. [A-Q ; B-P ; C-S ; D-R]

MULTIPLE CORRECT CHOICE TYPE

1. a, b 2. a, c 3. a, b 4. a, c 5. a, b 6. b, c

7. a, c 8. a, c, d

ASSERTION-REASON TYPE

1. D 2. A 3. B 4. C 5. D

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INITIAL STEP EXERCISE

(SUBJECTIVE)

1. 2.68 × 10–3 mol of a solution containing an ion An+

require 1.61 × 10–3 mol of MnO4– for the oxidation

of An+ to AO3– in acid medium. What is the value of

n ?2. How much 1.00 M HCl should be mixed with what

volume of 0.250 M HCl in order to prepare 1.00 Lof 0.500 M HCl ?

3. Calculate the final concentration of HNO3 if

0.20 mol HNO3 is added to a beaker containing

2.0 L of 1.1 M HNO3 and enough pure water is

added to give a final volume of 3.0 L.4. If 40.00 mL of 1.600 M HCl and 60.00 mL of

2.000 M NaOH are mixed, what are the molarconcentrations of Na+, Cl–, and OH– in theresulting solution ? Assume a total volume of100.00 mL.

5. Calculate the molarity of the original H3PO

4

solution if 20.0 mL of H3PO

4 solution is required to

completely neutralize 40.0 mL of 0.0500 M Ba(OH)2

solution.6. What volume of 96.0% H

2SO

4 solution (density 1.83

g/mL) is required to prepare 2.00 L of 3.00 M H2SO

4

solution ?7. How many mL of 0.5000 M KMnO

4 solution will

react completely with 20.00 g of K2C

2O

4.H

2O

according to the following equation ?16H+ + 2MnO

4– + 5C

2O

42– 10CO

2 + 2Mn2+ + 8H

2O

8. When 50.00 mL of a nitric acid solution was titratedwith 0.334 M NaOH, it requires 42.80 mL of thebase to achieve the equivalence point. What is themolarity of the nitric acid solution ? What mass ofHNO

3 was dissolved in 90.00 mL of solution ?

9. The acidic substance in vinegar is acetic acidCH

3COOH. When 6.00 g of a certain vinegar was

titrated with 0.100 M NaOH, 40.11 mL of base hadto be added to reach the equivalence point. Whatpercent by mass of this sample of vinegar is aceticacid ?

10. Calculate the percent of BaO in 29.0 g of a mixtureof BaO and CaO which just reacts with 100.8 mLof 6.00 M HCl. BaO + 2HCl BaCl

2 + H

2O ;

CaO + 2HCl CaCl2 + H

2O

11. Calculate the normality of each of the followingsolutions : (a) 7.88 g of HNO

3 per L solution (b)

26.5 g of Na2CO

3 per L solution (if acidified to form

CO2).

12. What volumes of 12.0 N and 3.00 N HCl must bemixed to give 1.00 L of 6.00 N HCl ?

13. One gram of a mixture of CaCO3 and MgCO

3 gives

240 ml of CO2 at N.T.P. Calculate the percentage

composition of mixture (Ca = 40, Mg = 24, C = 12,O = 16).

14. (a) What volume of 5.00 N H2SO

4 is required

to neutralize a solution containing2.50 g NaOH ?

(b) How many g pure H2SO

4 are required ?

15. A 0.250 g sample of a solid acid was dissolved inwater and exactly neutralized by 40.0 mL of 0.125N base. What is the equivalent weight of the acid ?

16. Exactly 50.0 mL of Na2CO

3 solution is equivalent

to 56.3 mL of 0.102 N HCl in an acid-baseneutralization. How many g CaCO

3 would be

precipitated if an excess of CaCl2 solution were

added to 100 mL of this Na2CO

3 solution ?

17. How many cm3 of concentrated sulfuric acid, ofdensity 1.84 g/cm3 and containing 98.0% H

2SO

4 by

weight, should be taken to make 1.00 L of normalsolution. Assume complete ionisation.

18. A 40.8 mL sample of an acid is equivalent to50.0 mL of Na

2CO

3 solution, 25.0 mL of which is

equivalent to 23.8 mL of a 0.102 N HCl. What is thenormality of the first acid ?

19. Given the unbalanced equationKMnO

4 + KI + (H)

2SO

4 (K)

2SO

4 + MnSO

4 + I

2 +

H2O

(a) How many g KMnO4 are needed to make

500 mL 0.250 N solution ?(b) How many g KI are needed to make

25.0 mL 0.360 N solution ?20. A solution contains 4 g of Na

2CO

3 and NaCl in

250 ml. 25 ml of this solution required 50 ml ofN/10 HCl for complete neutralization. Calculate %composition of mixture.

21. The density of a 2.0 M solution of acetic acid(MW = 60) in water is 1.02g/mL. Calculate the molefraction of acetic acid.

22. The density of a 2.03 M solution of acetic acid inwater is 1.017 g/mL. Calculate the molality of thesolution.

23. Calculate the (a) molar concentration and (b)molality of a sulfuric acid solution of density1.198 g/cm3, containing 27.0% H

2SO

4 by weight.

24. A solution contains 57.5 mL ethyl alcohol (C2H

5OH)

and 600 mL benzene (C6H

6). How many g alcohol

are in 1000 g benzene ? What is the molality of thesolution ? Density of C

2H

5OH is 0.800 g/mL; of

C6H

6, 0.900 g/mL.

25. One gram of an alloy of aluminium andmagnesium when treated with excess of dil. HClfroms magnesium chloride, aluminium chloride andhydrogen. The evolved hydrogen collected overmercury at 00 C has a volume of 1.20 litres at0.92 atm. pressure. Calculate the composition of thealloy. [H = 1, Mg = 24, Al = 27].

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26. A gas mixture of 3.0 litres of propane and butaneon complete combustion at 250C produced 10 litresof CO

2. Find out the composition of the gas

mixture.27. In an Industrial process for producing acetic acid,

oxygen gas is bubbled into acetaldehyde CH3CHO,

containing manganese (II) acetate (catalyst) underpressure at 600C.2CH

3CHO(l) + O

2(g) 2CH

3COOH(l)

In a laboratory test of this reaction, 20.0 g CH3CHO

and 10.0 g O2 were put into a reaction vessel. (a)

How many grams of acetic acid can be producedby this reaction from these amounts of reactants ?(b) How many grams of the excess reactantremaining after the reaction is complete ?

28. The hourly energy requirements of an astronautcan be satisfied by the energy released when34 grams of sucrose are “burned” in his body. Howmany grams of oxygen would he needed to becarried in spaces capsule to meet his requirementfor one day ?

29. 1.84 g of a mixture of CaCO3 and MgCO

3 are heated

strongly till no further loss of weight takes place.The residue weighs 0.96 g. Find the percentagecomposition of the mixture.(Mg = 24, Ca = 40, C = 12, O = 16)

30. 1.0 g of a mixture of Potassium chloride andPotassium iodide dissolved in water andprecipitated with Silver nitrate, gave 1.618 g ofsilver halides. Calculate the percentage of each inthe mixture (K = 39, Cl = 35.5, Ag = 108, I = 127).

31. 20% surface sites have adsorbed N2. On heating N

2

gas evolved from sites and were collected at0.001 atm and 298 K in a container of volume is2.46 cm3. Density of surface sites is6.023 × 1014/cm2 and surface area is 1000 cm2, findout the number of surface sites occupied permolecule of N

2.

32. A plant virus is found to consist of uniformcylindrical particles of 150 Å in diameter and5000 Å long. The specific volume of the virus is0.75 cm3/g. If the virus is considered to be a singleparticle, find its molar mass.

33. (a) Calculate the amount of calcium oxiderequired when it reacts with 852 g ofP

4O

10.

243104 )PO(Ca2OPCaO6

[At. wt. : Ca = 40, P = 31, O = 16](b) How many milliliters of 0.5 M H

2SO

4 are

needed to dissolve 0.5 g of copper (II)carbonate ?

224423 COOHCuSOSOHCuCO

34. 1.5 gm of an impure sample of (NH4)

2SO

4 was boiled

with excess of Caustic soda solution in a Kjeldahl’sflask and the ammonia evolved was passed into 200ml of semi-normal H

2SO

4 solution. Thepartially

neutralized acid was made to 500 ml with distilledwater. 25ml of this diluted acid required 40.8 ml ofdecinormal Caustic soda for completeneutralization. Calculate the percentage purity ofAmmonium sulphate.

35. A sample of a metal (M) carbonate was neutralizedby 10 ml of 0.1 N-hydrochloric acid and theresulting chloride gave 0.0517 g of phosphate[M

3(PO

4)

2]. Calculate the eq. wt. of M. (the formula

of Phosphoric acid is H3PO

4). Determine the atomic

weight of M ?

FINAL STEP EXERCISE

(SUBJECTIVE)

1. One litre of a mixture of O2 and O

3 at STP was

allowed to react with excess of acidified solution ofKI. The iodine liberated require 40mL of M/10sodium thiosulfate solution for titration. What isthe mass percentage of ozone in mixture.

222

KI3 OOIO

2. What volume of 3.00 M HNO3 can react completely

with 15.0 g of a brass (90.0% Cu, 10.0% Zn)according to the following equations ?Cu + 4H+(aq) + 2NO

3–(aq) 2NO

2(g) + Cu2+ + 2H

2O

4Zn + 10H+(aq) + NO3–(aq) NH

2+ + 4Zn2+ + 3H

2O

What volume of NO2 gas at 250C and 1.00 atm

pressure would be produced ?

3. A mixture of FeO and Fe3O

4 when heated in air to a

constant weight gains 5 % in its weight. Find thecomposition of initial mixture. (Fe = 56, O = 16)

4. A 10.0 g sample of “gas liquor” is boiled with anexcess of NaOH, and the resulting ammonia ispassed into 60 cm3 if 0.90 N H

2SO

4. Exactly 10.0

cm3 of 0.40 N NaOH is required to neutralize theexcess sulfuric acid (not neutralized by the NH

3).

Determine the percent ammonia in the “gas liquor”examined.

5. A 10.0 mL portion of (NH4)

2SO

4 solution was treated

with excess NaOH. The NH3 gas evolved was

absorbed in 50.00 mL of 0.1000 N HCl. Toneutralize the remaining HCl, 21.50 mL of 0.0980N NaOH was required. What is the molarconcentration of the (NH

4)

2SO

4 ? How many g

(NH4)

2 SO

4 are in 1 L solution ?

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6. A solution contained the mixture of Na2CO

3 and

NaCl. 25 ml of the solution required 20.4 ml of 0.095N-HCl to convert the carbonate into the chloride.25 ml of the later solution (containing chlorideconverted from carbonate) required 38.76 ml ofN/10-AgNO

3 to ppt, the chloride completely.

Calculate the strength of Na2CO

3 and NaCl in one

litre of the original solution.7. 10 g of a mixture of Cu

2S and CuS was treated with

200 mL of 0.75 M MnO4

– in acidic solutionproducting SO

2, Cu2+ and Mn2+ . SO

2 was bubbled

off and excess of MnO4– was titrated with 175 mL

of 1.00 M Fe2+. Calculate the CuS in the mixture.8. Determine the volume of dilute nitric acid (density

1.11 g/mL, 19.0% HNO3 by weight) that can be

prepared by diluting with water 50 mL of theconcentrated acid (density 1.42 g/mL, 69.8% HNO

3

by weight). Calculate the molar concentrations andmolalities of the concentrated and dilute acids.

9. A 1.00 g sample of KClO3 was heated under such

conditions that a part of it decomposed accordingto the equation(i) 2KClO

3 = 2KCl + 3O

2

and the remaining underwent change according tothe equation(ii) 4KClO

3 = 2KClO

4 + KCl

If the amount of oxygen evolved was 146.8 ml atS.T.P., calculate percentage by weight of KClO

4 in

the residue (K = 39.1, Cl = 35.5)10. To a 25 ml H

2O

2 solution, excess of acidified

solution of potassium iodide was added. The iodineliberated required 20 ml of 0.3 N sodiumthiosulphate solution. Calculate the volume strengthof H

2O

2 solution.

11. A solution contains Na2CO

3 and NaHCO

3. 20 cm3

of this solution requires 5.0 cm3 of 0.1 M H2SO

4

solution for neutralization using phenophthalein asthe indicator. Methyl orange is then added andfurther 5.0 cm3 of 0.2 M H

2SO

4 was required.

Calculate the masses of Na2CO

3 and NaHCO

3 in

1 L of this solution.12. A 1.2 g of a mixture containing H

2C

2O

4.2H

2O and

KHC2O

4.H

2O and impurities of a neutral salt,

consumed 18.9 ml of 0.5 N NaOH forneutralization. On titrating with KMnO

4 solution

0.4 g of the same substance needed 21.55 ml of 0.25N KMnO

4. Calculate the percentage composition

of the substance.13. A mixture of FeO and Fe

2O

3 is reacted with

acidified KMnO4 solution having a concentration

of 0.2278 M, 100 ml of which was used. Thesolution then was treated with Zn dust whichconverted the Fe3+ of the solution to Fe2+ . The Fe2+

required 1000 ml of 0.13 M K2Cr

2O

7 solution. Find

the mass % of FeO and Fe2O

3.

14. 10 g of a mixture of anhydrous nitrates of two met-als A and B were heated to a constant weight andgave 5.531 g of a mixture of the corresponding ox-ides. The equivalent weights of A and B are 103.6and 31.8 respectively. What was the percentage ofA in the mixture ? (N = 14, O = 16).

15. Equal weights of mercury and iodine are allowedto react completely to form a mixture of mercurousand mercuric iodides. Calculate the ratio of theweights of mercurous and mercuric iodides formed.(I = 127, Hg = 201).

16. 5 ml of 8 N nitric acid, 4.8 ml of 5 N HCl acid and acertain volume of 17 M sulphuric acid are mixedtogether and made up to 2 litre. Thirty ml of thisacid mixture exactly neutralise 42.9 ml of sodiumcarbonate solution containing one gram of Na

2CO

3.

10H2O in 100 ml of water. Calculate the mass in

gram of the sulphate ions in solution.17. 200 ml of a solution of mixture of NaOH and

Na2CO

3 was first titrated with phenolphthalein and

N/10 HCl. 17.5 ml of HCl was required for the endpoint. After this methyl orange was added and 2.5ml of same HCl was again required for next endpoint. Find out amounts of NaOH and Na

2CO

3 in

mixture.18. 50 ml of solution, containing 1g each of Na

2CO

3,

NaHCO3 and NaOH, was titrated with N HCl.

What will be the titre readings if(a) only phenolphthalien is used as

indicator ?(b) only methyl orange is used as indicator

from the very beginning ?(c) methyl orange is added after the first end

point with phenolpthalein ?19. A mixture of calcium carbonate and sodium

chloride weighing 3.20 g was added to 100 ml of1.02 N HCl. After the reaction had ceased theliquid was filtered, the residue washed and thefiltrate was made up to 200 ml. 20 ml of this dilutesolution required 25 ml of N/5 NaOH forneutralization. Calculate the % of CaCO

3 in the

mixture. (Ca = 40).20. A solution contains a mixture of sulphuric acid and

oxalic acid, 25 ml of the solution require 35.5 ml ofN/10-NaOH for neutralization and 23.45 ml ofN/10 - KMnO

4 for oxidation. Calculate (a) the

Normality of the solution with regard to sulphuricacid and oxalic acid (b) the number of g of each ofthese substances present in one litre of the solution.

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ANSWERS SUBJECTIVE (INITIAL STEP EXERCISE)

1. n = 2 2. 333 mL of 1m HCl 3. 0.80 m4. 0.56 M OH–, 0.640 Cl–, 1.2MNa+ 5. 0.0667 M6. 335 mL 7. 86.9 mL 8. 1.62 gm, 0.286 M9. 4.01 % 10. 65.5 % BaO11. (a) 0.1251 N (b) 0.5 N 12. .33 L, .667 L13. CaCO

3 = 62.5 %, MgCO

3 = 37.5%

14. (a) 12.5 mL (b) 3.07 gm15. 50g/eq 16. 0.575 gm 17. 27.2 cm3

18. 0.119 N 19. (a) 3.95 gm (b) 1.49 gm20. Na

2CO

3 = 66.25%, NaCl = 33.75% 21. 0.038

22. 2.27 m23. (a) 3.30 m (b) 3.78 m 24. 0.54 kg C

6H

6, 1.85 m

25. Al% = 54.87%, Mg% = 45.13%26. 66.66% propane, 33.33 % butane 27. (a) 27.24 gm

(b) 2.73 gm28. 916.2 gm29. CaCO

3 = 54.35%, MgCO

3 = 45.65 % 30. KCl = 39.6 %

KI = 60.4%31. 2 32. 70.96 × 106 g/mol 33. (a) 1008 g (b) 8.097 mL34. 80.96 35. 20.03, 40.06

ANSWERS SUBJECTIVE (FINAL STEP EXERCISE)

1. 6.57 % 2. L4.10V,L302V23 NOHNO

3. % of FeO = 20.25%

% of Fe3O

4 = 79.75 % 4. 8.5% 5. 0.145 M, 19.1 g/L

6. Na2CO

3 = 4.109 gm, NaCl = 4.535 gm 7. CuS = 57.4 %

8. 15.7 M, 3.35 M, 36.8 m, 3.73 m 9. 49.83%

10. 1.344 11. Na2CO

3 = 5.3 gm, NaHCO

3 = 4.2 gm

12. KHC2O

4.H

2O = 80.9 %, H

2C

2O

4.2H

2O = 14.7 %

13. % FeO = 13.34 %, % Fe2O

3 = 86.66 % 14. 32.28

15. 0.513 : 1 16. 6.528 gm 17. WNaOH

= 0.06gm

gm0265.0W32CONa

18. (a) 34.4 mL (b) 55.8 mL (c) 21.3 mL

19. 81.25% 20. H2SO

4 = 0.0482N, 2.36 g/L

H2C

2O

4 = 0.0938N, 4.225 g/L

Documents

CS– 1 STOICHIOMETRY - Einstein Classeseinsteinclasses.com/Stoichiometry.pdf · STOICHIOMETRY C1 In this chapter ... In aq. solution n = MV[M - molarity, V ... It is used in acid