STOICHIOMETRY 

 

Hints for Subjective Problems

 

LEVEL – I

 

5. ⎯→ KCl + O2 (Part – I)

⎯→ + KCl (Part – II)

To calculate (x) apply POAC for ‘O’ atom in Part (I)

For calculation of y apply POAC for ‘O’ atom in Part (II) and calculate % of KClO4 in the residue as follow.

% of KCl4 =

7. Let millimoles of Fe3O4 and Fe2O3 in the mixture is x and y respectively. As Fe3O4 and Fe2+ hence n-factor for both is 2.

∴ Meq. of Fe3O4 + Meq. of Fe2O3 = Meq. of Na2S2O3

⇒ 2x + 2y = 7.2 × 1 × …(1)

In second redox titration

Fe2+ ⎯→ Fe3+ n = 1

MnO4– ⎯→ Mn2+ n = 5

For Fe2+, n = 1, ∴ For Fe2+ millimoles = milli equivalent

∴ Millimoles of Fe2+ from Fe3O4 + Millimole of Fe2+ for 

Fe3O4 = Meq. of KMnO4

∴ 3x + 2y = 4.2 × 5 × …(2)

From equation (1) and (2) calculate x and y and proceed further

 

LEVEL – II

 

1. For FeSO4, Fe2+Fe2+
2. O3 + 2KI + H2O ⎯→ 2KOH + I2 + O2

Redox changes are I2 + 2e ⎯→ 2I–

2S22+ ⎯→ + 2e

8. Meq. of Na2HAsO3 = Meq. of I2 in first case …(1)

Meq. of Na2HAsO3 + Meq. of As2O5 = Meq. of Na2S2O3 second case …(2)

10. Li2CO3 LiO + CO2

Na2CO3 No change

Li2CO3 + Na2CO3 LiCl + NaCl + CO2 + H2O

Meq. of Li2CO3 + Meq. of Na2CO3 = Meq. of HCl

 

LEVEL – III

1. An 1st step Meq. of NaHSO3 = Meq. of NaIO3 =

(n- factor = 6)

In second step n-factor

For I– is 1 and n-factor for IO3– is 5

IO3–  + 5I– + 6H+⎯→ 3I2 + H2O

2. As balanced equation is given it is advisable to solve problem by taking mole ratio
3. i) As ClO3– is reduced to Cl–.

Hence moles of AgCl is due to Cl– from ClO3– and from KCl

1. ii) of ClO3– in 25 ml solution 

= Meq. of FeSO4 taken – Meq. of Fe2SO4 left.

= Meq. of FeSO4 reacted

 

Subjective Problems

 

LEVEL – I

 

1. NH4Cl reacts with NaOH. The excess of NaOH is utilized by H2SO4.

Meq. of NaOH initially taken = 100 × 0.8 = 80

Meq. of NaOH left after reaction = Meq. of H2SO4 = 12.5 × 0.75 = 9.375

∴ Meq. of NaOH used for NH4Cl = 80 – 9.375 = 70.625

∴ Meq. of NH4Cl = 70.625

× 1000 = 70.625

⇒ × 1000 = 70.625

∴ = 3.78 gm

2. Meq. of NaOH added = 30 × 0.04

Meq. of Alkali left = 22.48 × 0.024 = 0.54

Meq. of alkali used for SO2 and H2O2 = 0.66

Weight of alkali used = = 0.0264

80 gm of NaOH reacts with 64 gm of SO2

∴ 0.0264 gm of NaOH reacts = = 0.021 gm SO2

∴ 64 gm SO2 required = 32 gm S

∴ 0.021 gm SO2 required = = 0.0105 gm

∴ % of S = ×1000 = 6.1875 %

3. NHypo× 12.3 = 24.6 × 0.5

∴ NHypo = 1

2KClO3 + 12NHCl ⎯→ 2KCl + 6H2O + 6Cl2

Cl2 + KI ⎯→ 2KCl + I2

Meq. of I2 = Meq. of Cl2 = Meq. of hypo = 100 × 1 = 100

∴ m mole of Cl2 = = 50

⇒

∴ = 2.042

% of KClO3 = × 100 = 82.32%

4. Meq. of 25 ml H2O2 = meq. of I2 = meq. of Na2S2O3 = 20 × 0.3 = 6

∴ Normality of H2O2 =

Let volume strength of H2O2 be ‘xv’. Thus 1 litre of H2O2 produces x litre of O2 at NTP.

∴ Equivalent of H2O2/lit = Eq. of O2

=

∴ ∴ x = 1.344

5. KClO3⎯→ KCl + O2

Applying POAC for O atoms in equation (1)

3 × moles of KClO3 = 2 ×moles of O2

⇒ 3 ×

⇒ = 0.5358 gm

Applying POAC for K atoms

1 × moles of KClO3 = 1 × moles of KCl

⇒ ⇒ × 74.5 = 0.3260 gm

In second reaction amount of KClO3 = 1 – 0.5358 = 0.4642 gm

KClO3 ⎯→ KClO4 + KCl

0.4642 gm

Applying POAC for O atoms

3 × moles of KClO3 = 4 × moles of KClO4

⇒

⇒ = 0.3937 gm

Weight of KCl produced by second reaction

= Weight of KClO3 – Weight of KClO4

= 0.4642 – 0.3937  0.0705 gm

Now, since on heating KClO3, O2  will escape out, the substances as residue are KCl produced by reactions (i) of (ii) and KClO4.

Weight of residue = 0.3260 + 0.3937 + 0.0705 = 0.7902 gm

∴ % of KClO4 in residue = × 100 = 49.8%

6. The loss in weight (28% of 5 gm i.e. 1.4 gm is due to the formation of gases NO2+ and O2 which escape out?

Pb(NO3)2 ⎯⎯ PbO    +   NO2 + O2

x gm (say)       (x – y)       y gm (say)

NaNO3 ⎯→ NaNO2         +     O2

(5 – x)          (3.6 – x +y)         (1.4 – y)

Applying POAC for Pb and Na atoms, we get respectively.

Mole of Pb(NO3)2 = mol of PbO

⇒ …(1) { = 331, = 223}

Mole of NaNO3 = Mole of NaNO2

⇒ …(2)

From equation (1) and (2)

x = 3.3246 = amount of Pb(NO3)2

Amount of NaNO3 = 5 – 3.3246 = 1.675 gm

7. This problem can be done by two methods.

In the first method, we break up Fe3O4 as an equimolar mixture of FeO and Fe2O3.

Method (1)

Fe3O4 is FeO. Fe2O3

Equivalents of Na2S2O3 = =   7.2 × 10-3

Equivalents of I2 in 10 cc = 7.2 × 10-3

Equivalents of I2 in 50 cc = 7.2 × 10-3 × 5 =   3.6 × 10-2

Since equivalents of I2 is equal to that of KI which in turn is equal to the total equivalents of Fe2O3 (Fe2O3 in the free state and Fe2O3 combined with FeO).

∴ equivalents of total Fe2O3     = 3.6 × 10-2

Equivalents of KMnO4 solution   = = 2.1 × 10-2

Since KMnO4 reacts with the total Fe2+ (Fe2+ in FeO and Fe2+ that was produced by the action of KI on Fe2O3) 

∴ equivalents of total Fe+2 in 50 mL = 2.1 × 10-2 × 2 = 4.2 × 10-2 

Since equivalents of Fe2+ produced from Fe2O3 is equal to the equivalents of Fe2O3 

∴ Equivalents of FeO = 4.2 × 10-2 – 3.6 × 10-2

= 6 × 10-3

∴ moles of FeO = 6 × 10-3

moles of Fe2O3 combined with FeO = 6 × 10-3

total moles of Fe2O3 =
(because when Fe2O3 → Fe2+ ‘n’ factor is 2).

= 1.8 × 10-2

moles of Fe2O3 in the free state  =  1.8 × 10−2  –  6 × 10−3

=  1.2 × 10-2

mass of Fe3O4 = 6 × 10-3 × 232  = 1.392 g

mass of Fe2O3 =  1.2 × 10-2 × 160 = 1.92 g

percentage Fe3O4 =  17.4%

percentage of Fe2O3 =  24%

Method 2:

Here we take Fe3O4 as a single entity.

Equivalents of Na2S2O3 =   = 7.2 × 10-3

Equivalents of I2 in 50 cc = 7.2 × 10-3 × 5 = 3.6 × 10-2

∴ Equivalents of Fe3O4 + Fe2O3 = 3.6 × 10-2

Let us assume that the moles of Fe3O4 is x and that of Fe2O3 is y. Since on reacting with KI both Fe3O4 and Fe2O3 give Fe2+ ‘n’ factor for both is two.

∴ 2x + 2y = 3.6 × 10-2 ——–   (1)

Equivalents of KMnO4 = = 2.1×10-2

Equivalents of Fe2+ in 50 mL = 2.1 × 10-2 × 2 = 4.2 × 10-2

Moles of Fe2+ in 50 mL = 4.2 × 10-2

Since the moles of Fe3O4 are x, moles of Fe2+ produced from Fe3O4 will be 3x and that produced from Fe2O3 will be 2y. 

∴ 3x + 2y = 4.2 × 10-2  ——— (2)

(2) – (1) gives x = 6 × 10-3       ∴ y = 1.2 × 10-2

Solving this, percentage of Fe3O4 is 17.4% and Fe2O3 is 23.75%.

8. Equivalents of H2O2 intially = = 1.176

Equivalents of Sn2+ =

Equivalents of H2O2 left = 1.176 − 0.42 = 0.756

Moles of H2O2 left =

Moles of O2 produced = = 0.189

Volume of O2 = = 4.54 L

9. Equivalents of K2Cr2O7 =

Equivalents of Fe2+ in 50 ml = 1 × 10−3

Equivalents of Fe2+ in 250 ml = 1 × 10−3 × 5 = 5 × 10−3 

Initial equivalents of Fe2+ =

Equivalents of Fe2+ consumed by MnO2 = 2.55 × 10−2 − 5 × 10−3 = 0.0205

∴ Equivalents of MnO2 = 0.0205

Moles of MnO2 =

Mass of MnO2 = 1.025 × 10−2 × 87 = 1.0875g

% MnO2 = =  72.5%

10. Let the mass of H2C2O4.2H2O and KHC2O4.H2O be x and y gm. Respectively.

∴ × 2 + × 1 =

× 2 + × 2 =

∴ x = 0.1764 gm, y = 0.9709 gm.

% H2C2O4H2O = × 100 = 14.7%

% KHC2O4.H2O = 80.9%

 

LEVEL – II

1. For FeSO4 : Fe2+Fe3+ Fe2+

Meq. of Fe2+ in FeSO4 + Meq. of Fe2+ in FeC2O4  + Meq. of C2O42– = Meq. of KMnO4 used

∴x × 1 + y × 1 + y × 2 = 40 =  

(Let millimole of FeSO4 and FeC2O4 be x & y respectively)

∴ x + 3y = …(1)

Meq. of Fe2+ in FeSO4 + Meq. of Fe2+ in FeC2O4 = Meq. of KMnO4 used

(Because after reduction of mixture only Fe2+ ions formed from Fe3+. Since CO2 escapes out in air

∴ x ×1 + y × 1 = 25 ×

∴ x + y = …(2)

By equation (1) + (2) x = , y =

∴ =

2. The redox changes are as follows

For FeCl3 Fe3+ + e ⎯→ Fe2+

For N2H6SO4 N22– ⎯→ N20 + 4e

For KMnO4 Mn7+ + 5e ⎯→ Mn2+

Meq. of N2H6SO4 in 10 ml solution = Meq. of FeCl3 reacting with N2H6SO4  = Meq. of Fe2+ formed = Meq. of KMnO4

∴ Meq. of N2H6SO4 in 10 ml = 20 × × 5 = 2

∴ × 4 × 1000 = 2

∴ W = 0.065

Weight of N2H6SO4 in 1000 ml = = 6.5 gm/lit

3. H2O + 2KI + O3 ⎯→ 2KOH + I2 + O2

Redox changes are

I2 + 2e ⎯→ 2I–

2S22+ ⎯→ S4+5/2 + 2e

(I2 + 2Na2S2O3⎯→ 2NaI + Na2S4O6)

Meq. of I2 = Meq. of Na2S2O3

= 1.5 × 0.01 = 1.5 × 10–2

= Millimoles of I2 = = 7.5 × 10–3

Millimoles of O3= Milli mole of I2 = 7.5 × 10–3

∴

= 184.725 ×10–7 atm

Volume % of O3 = × 100

= 1.847 × 10–3%

4. Let two carbonates are M′2CO3 and M″2CO3 since alkali metals are monovalent

Also M′2CO3 = a gm, M″2CO3 = b gm

a + b = 1 …(1)

= 2 × 7 + 60 = 74, = 2m + 60

∴ = …(2)

Meq. of carbonate 1 + Meq. of carbonate (2)  = Meq. of HCl

∴ × 1000 + = 44.4 × 0.5 …(3)

Solving equation (1), (2) and (3)

m = 23, a = 0.41 gm, b = 0.59 gm

Meq. of sulphates (M′2SO4 = Meq. of M′2CO3) = × 1000 = 11.08

= = 11.08, ∴ = 0.609 gm

Meq. of sulphate (M′2SO4) = Meq. of M″2CO3

∴ × 1000 = × 1000

∴ = 0.7902 gm

∴ Weight of sulphates = 0.6094 + 0.7902 = 1.3996 gm         

5. Let Weight of Na2CO3 = a gm

Weight of K2CO3 = b gm

a + b = 1 …(1)

Solution reacts with HCl to give chlorides which further react with AgNO3 

For 250 ml solution, (Meq. of Na2CO3 + Meq. of K2CO3) = Meq. of HCl = Meq. of AgNO3.

∴ + = 20 × N ×  

= 16.24 × 0.1 × …(2)

NHCl = 0.0812 = MHCl

Strength of HCl = 0.0812 × 36.5 gm/lit = 2.96gm/lt

From equation (1) and (2)

a = 0.4 gm

b = 0.6 gm

∴ % of K2CO3 = 60%

6. The first point that is to be made clear here is about % (by mole) yield. If a reaction occurs with y% (by mole) yield, then it implies that the moles of product produced would be y% of the moles that would have been produced had the reaction gone with 100% (by mole) yield. Now let us assume that x L of C2H6 at STP would produce the desired amount of C4H10, then the moles of C2H6 would be =

The moles of C2H5Br that would have been produced with 100% (by mole) yield = [ C2H5Br and C2H6 are in the molar ratio of 1:1]

But the moles of C2H5Br actually produced = 

Moles of C4H10 that would have been produced with 100% (by mole) yield

= [C2H5Br and C4H10 are in the molar ratio of 2:1]

But the moles of C4H10 that would be actually produced 

=

∴ The mass of C4H10 produced =

This must be equal to 55 g.

∴

x  =  55.5 L

7. The First step towards solving any problem is to read the problem over and over again till you understand what is given and what is asked. In this problem we have to calculate a certain volume of AgNO3 solution that can precipitate all the chloride ions in any possible mixture with any composition.

This implies that we must find that combination which contains the maximum Cl− ions and find the volume of AgNO3 corresponding to that. This is because if this much volume of AgNO3 can precipitate all Cl– ions from the sample containing maximum chloride ions then it can precipitate all the Cl− ions from any other combination. Now the only confusion you may have is that when minimum volume is asked then why calculate the volume of AgNO3 solution for maximum Cl– ion. This would become clear in a moment.

Let us find out that combination that gives us the maximum moles of Cl−.

If we take pure cases, 0.3 g of each of the following has the following moles.

0.3 g of NaCl = 

0.3 g of NH4Cl =

0.3 g of KCl =  moles

So out of the pure cases, it is pure NH4Cl which has the maximum moles of Cl− ions. Let us take 0.3 g of NH4Cl and replace some of it with NaCl or KCl. If we take x gm of NH4Cl out and replace it with NaCl or KCl then we would be taking out moles of NH4Cl and replacing it with or moles of NaCl or KCl respectively. This clearly indicates that any contamination of the NH4Cl mixture would lead to a decrease in the moles of Cl−. Therefore it is pure NH4Cl which has the maximum moles of Cl−. For calculation of the volume of AgNO3 solution all we need to do is to equate the moles of AgNO3 with the moles of Cl− [when AgNO3 reacts with Cl− to give AgCl, it would react in a molar ratio of 1:1]

 

Moles of AgNO3 in volume Vml = V × 1.04 ×

[Note: Specific gravity is the ratio of density of a solution to the density of water at 4oC. Since the density of water at 4oC is 1 gm/cc,  the sp. gravity of any solution also becomes its density in gm/cc].

Moles of Cl− in pure 0.3 g of NH4Cl =

∴ = V×1.04 ×

∴ V = 18.33 ml

Now, let us try to understand why we call it the minimum volume. It is clear that 18.33 ml of AgNO3 solution can precipitate all the Cl− from 0.3 g of NH4Cl and since this is the maximum Cl− content, this much volume of AgNO3 solution has sufficient Ag+ ions to precipitate all the Cl− from any other combination also. It can also be seen that any volume V such that V ≥ 18.33 ml can precipitate  all the Cl− ions from any combination. Therefore the least volume that can be used to precipitate all the Cl− ions is 18.33 ml.

8. Equivalents of I2 reacting = = 3.39 × 10–3

∴ Equivalents of Na2HAsO3 = 3.39 × 10–3

Moles of Na2HAsO3 = = 1.695 × 10–3

Mass of Na2HAsO3 = 1.695 × 10–3 × 170 = 0.289 gm

% Na2HAsO3 = 11.6%

Equivalents of Na2S2O3 =  = 4.944 × 10–3

∴ Equivalents of I2 liberated = 4.994 × 10–3

∴ Equivalents of As2O5 and Na2HAsO3 = 4.994 × 10–3

∴ Equivalents of As2O5 = 4.994 × 10–3 – 3.39 × 10–3 = 1.554 × 10–3

Moles or As2O5 = = 3.885 × 10–4

ass of As2O5 = 3.885 × 10–4 × 230 = 0.0893 gm 

% As2O5 = × 100 = 3.57%

9. I2 + H2S ⎯→ 2HI + S ↓

250 ml diluted  tincture of iodine solution has  

equivalents of I– of n factor 2.

[Note: In the first reaction n factor of I– produced is 1 but when it is reacting with Ce4+ to give ICl its n factor is 2 ∴ equivalents of I– in the second reaction with n factor 2 to be converted into equivalents of I–  of n factor 1.].

∴ equivalents of I– reacted of n–factor 1 in 250 ml =

Equivalents of I– produced ion 1 litre= = equivalents of I– produced
in 5 ml.

= moles of I– produced in 5 ml.

∴ weight of entire I– = g 

1n  5 ml = 0.3556 g 

2 ml tincture of iodine gives 

0.0313 g AgBr (yellow ppt) this means that weight of I– in 2 ml tincture of iodine 

= = 0.0169 g 

∴ In 5 ml tincture of iodine 0.0169 ×2.5 = 0.042 g I–

∴ % weight of Iodine in the form of free iodine  = ×100 

= 88.2%

10. Out of all alkali metal carbonates, its only Li2CO3 which decomposes to
give oxide and CO2, rest all alkali metal carbonates are stable
to thermal decomposition.    

Li2CO3 Li2O + CO2

1. of moles  of Li2CO3 =  no.of moles  of CO2

=  = 1.226 × 10–3

In the residue Li2O, Na2CO3 and Na2O will be present when they react with HCl, CO2 will be released by Na2CO3 only

∴moles  of Na2CO3 = moles  of CO2 evolved

= = 1.781 × 10–3

And equivalents of Na2CO3+equivalents of Na2O + equivalents of Li2O = equivalents of HCl used. (Na2O  and Li2O react with HCl to form the respective chlorides and water, because they are basic).  

1.781 × 10–3 × 2  +  equivalents of Na2O +  1.226 × 10–3 × 2  = 15 × × 10–3

Equivalent of Na2O = 7.5 × 10–3 – 6.01 × 103 = 1.49 × 10–3

Equivalent  weight of Li2CO3 = 73.8/2 = 36.94

Equivalent weight of Na2CO3 = 106/2 = 53

Equivalent  weight of Na2O = 62/2 = 31

Weight of Li2 CO3 = 2.45 × 10–3 × 36.94 = 0.0905 gm

Weight of Na2CO3 = 3.56 × 10–3 × 53 = 0.1886 gm

Weight  of Na2O = 1.49 × 10–3 × 31 = 0.0462

% of Na2O in the mixture = = = 14.2%

LEVEL – III

1. Meq. of NaHSO3 = Meq of NaIO3 = N × V =

I5++ 6e ⎯→ I–  (For KIO3, n-factor = 6)

∴ Meq. of NaHSO3  = 175.76

× 1000 = 175.76 ⇒ = 9.14 gm

Also Meq. of I– formed using n-factor 6 = 175.76

In step II n-factor of I– is 1 and n-factor for IO3– is 5

Meq. of I– formed using n-factor 1 =

Meq. of NaIO3used in 2nd step =

⇒ ∴ = 20 mL

2. AgNO3 + KI ⎯→ AgI↓ + KNO3

Ag present in AgNO3 is removed as AgI by adding 50 mL of KI solution of which 20 mL requires 30 mL of M/10 KIO3

Meq. of KI in 20 mL = Meq. of KIO3 = 30 × × 4 For KIO3

∴ Meq. of KI in 50 ml added to AgNO3 I5+ + 4e →I+

= = 30 I– → I–1 + 2e

Meq. of KI left unused by AgNO3 = 50 × × 4 = 20 ∴ Eq. wt. of KI =

∴ Meq. of KI used for AgNO3 = 30 – 20 = 10

∴ Meq. of AgNO3 = 10

× 2 × 1000 = 10, ∴ = 0.85 gm = 85%

∴ If equivalent weight of KI =

Equivalent weight of AgNO3 =

3. 1 mL of Na2S2O3 = 0.0499 gm of CuSO4.5H2O = = 2 × 10–4 equivalent of CuSO4.5H2O

2Na2S2O3 + I2 ⎯→ 2NaI + Na2S4O6

Mol of Hg5(106)2 = = 5 × 10–4

I2 formed from 5 × 10–4 mol of Hg5(IO6)2 = 8 × 5 × 10–4 = 40 × 10–4 mol

Mol of Na2S2O3 required = 80 × 10–4mol

Volume (in mL) of Na2S2O3 required = = 40 mL

4. Let molecular weight of oxalate salt is M
5. i) n-factor in acid base reaction = 2
6. ii) n-factor in redox titration = 2 × Z                (C2O42– ⎯→ 2CO2 + 2e)

∴ Meq. of acid in 30 ml = Meq. of NaOH used

30 × = 27 × 0.12 …(1)

Also 30 × = 36 × 0.12 …(2)

From equations (1) & (2) = …(3)

Also total cationic charge = total anionic charge

∴ x + y = 2Z …(4)

By equations (3) & (4)

x : y : z :: 1 : 3 : 2

These are in simplest ratio and molecular formula is KH3(C2O4)2.nH2O

Molecular weight of salt = 39 + 3 + 176 + 8n = 218 + 18n

From equation (1) M = = 254.16

∴ 218 + 18n = 254.15 ∴ n = 2

∴ Oxalate salt is KH3(C2O4)2.2H2O

5. Let total moles of Mg used for MgO and Mg3N2 be a and b respectively

2Mg + O2 ⎯→ 2MgO

a 0

0 a

3Mg + N2 ⎯→ Mg3N2

b 0

0 b/3

Now moles of MgO and Mg3N2 are present in the mixture

MgO + 2HCl ⎯→ MgCl2 + H2O

Mg3N2 + 8HCl ⎯→ 3MgCl2 + 2NH4Cl

or solution contains (a) mol of MgCl2 from MgO and (b) moles of MgCl2  from Mg3N2 and moles of NH4Cl.

Now moles of HCl or Meq. of HCl (monobasic) = 60 – 12 = 48

2a + = 48 …(1)

Moles of NH4Cl formed = Moles of NH3 liberated = moles of HCl used for absorbing NH3

= (10 – 6) = 4 …(2)

From (1) 2a + = 48 or a = 16

∴% of Mg used for Mg3N2 = × 100 = 27.27%

6. ClO3– is reduced to Cl– by SO2 hence AgCl is due to total Cl–

Also ClO3– is reduced to Cl– by Fe2+ 

Meq. of Fe2+ initially taken = 30 × 0.2 = 6

Meq. of Fe2+ unused = 37.5 × 0.08 = 3

∴ Meq. of Fe2+ used = 6.0 – 3.0 = 3.0

Thus Meq. of ClO3– in 25 ml = 3.0

Moles of ClO3– in 25 ml = = 0.0005

+5ClO3– ⎯→ –1Cl– (n-factor 6)     

0.N  5            –1

Thus moles of ClO3– 25 m solution = 0.0005

ClO3– is also reduced to Cl– by SO2

In first experiment and precipitated as AgCl

Thus Cl– formed from ClO3– = AgCl from ClO3– = 0.0005

Total AgCl formed both from actual and Cl– from ClO3– = 0.1435 gm 

= = 0.0010 mol

Thus AgCl due to Cl– only = 0.0010 – 0.0005 = 0.0005 mol

Thus ClO3– and Cl– are in molar ratio = 1:1

7. Cr (in steel) ⎯→

52 g           108g 253g

            0.0549 g

From the weight of given BaCrO4, we can determine weight of Cr present in 10 mL solution.

Weight of chromium = × 0.0549 g

= 0.0113 g in 10 mL portion

= 0.0113 × 25 in 250 mL solution

= 0.2821 g

This amount is in 10.00 g steel sample. Hence % of Cr in steel sample 

= = 2.821%

Fe2+ reduces as well as in acidic medium

5Fe2+ + + 8H+ → Mn2+ + 5Fe3+ + 4H2O

6Fe2+ + + 14H+ → 2Cr3+ + 6Fe3+ + H2O

This titration will give total normality of and

10 mL of this mixture solution = 15.95 mL of 0.075 M

= 15.95 mL of 0.075 N

∴ N( + ) = = 0.1196 N

From the weight of BaCrO4, we can also deduce weight of and hence is normality.

Weight of in 0.0549 g BaCrO4 = = 0.0234 g in 10 mL

equivalent weight of =

∴ N( = = = 0.065 N

thus N () = 0.0545 N by difference

equivalent weight of =

in250 mL solution = = 0.3243 g

Mn in 250 mL solution = 0.1498 g

This is present in 10.00 g steel sample

Hence % of Mn = = 1.498%

8. CuFeS2 + 2H2SO4 → CuSO4 + FeSO4 + 2H2S↑

Thus mineral is converted into CuSO4 and FeSO4

Titration – I: oxidises Fe2+ into Fe3+

+ 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O

10 mL of Fe2+ (in mineral) ≡ 2 mL of 0.01 M

≡ 2 mL of 0.05 N

10 × N (Fe2+) ≡ 2 × 0.05 N ()

∴ N (Fe2+) ≡ 0.01

∴ Fe2+ in mineral = 0.01 × 56

= 0.56 gL–1 present in 10 gL–1 solution

∴ % of Fe in mineral = = 5.6%

Titration II: Cu2+ is reduced to Cu+ (as white ppt. Cu2I2) on treatment with I–. I2 formed is titrated using

2Cu2+ + 4I– → Cu2I2 + I2

I2 + → + 2I–

This titration will give amount of Cu2+

25 mL of Cu2+ solution ≡ 5 mL of 0.01 M

≡ 5 mL of 0.01 N

∴ 25 × N (Cu2+) = 5 × 0.01 N

N (Cu2+) = 0.002

∴ Cu2+ in mineral = 0.002 × 63.5 gL–1

= 0.127 gL–1 present in 10 gL–1 solution

% of Cu = = 1.27%

thus Fe content in mineral = 5.6%

and Cu content is = 1.27%

9. Equivalents of Barium permanganate = = 1.485 ×10–4

Equivalents of Fe2+ in excess = 1.485 × 10–4

Initial equivalents of Fe2+ = = 8 × 10–3

Equivalents of Fe2+ consumed by Cr6+ = 8 × 10–3 – 1.485 × 10–4 = 7.85 × 10–3

∴Equivalents of Cr6+ = 7.85 × 10–3

Equivalents of FeO. Cr2O3 = 7.85 × 10–3

Moles of FeO. Cr2O3 = = 1.3 × 10–3

Mass of FeO. Cr2O3 = 1.3 × 10–3 × 224 = 0.2912 gm.

% FeO.Cr2O3 = × 100 = 15. 57%

10. Let weight of BaCO3, CaCO3 and CaO are

x, y and z respectively

∴ x + y + z = 1.249

BaCO3 ⎯→ BaCrO4

Redox charge Cr6+ + 3e ⎯→ Cr3+

2I– ⎯→ I2 + 2e

Meq. of BaCO3 = Meq. of BaCrO4 = Meq. of I2

∴ × 1000 = 20 × 0.05 ×

∴ x = 0.657 gm …(1)

The equivalent weight of BaCrO4 is M/3 therefore for BaCO3, it should be M/3 also because mole ratio of BaCO3 and BaCrO4 is 1 : 1

Applying POAc for C atom

Moles of C in BaCO3 + Moles of in CaCO3 = Moles of C in CO2

∴

⇒ 200x + 294y = 295.5 …(2)

From equation (1) and (2)

y = 0.416 gm

∴ 0.0657 + 0.416 + z = 1.249

z = 0.176

% of CaO = = 14.09%

11. CO2 gas is acidic in nature, it neutralizes the Ba(OH)2 and precipitates BaCO3.

Equations of Ba(OH)2 consumed = equivalents of CO2

equivalents of CO2 = – =

∴ moles of CO2 =

∴% of CO2 in air at NTP = = 0.0224 

Ba(OH)2 + CO2 ⎯→ BaCO3 + H2O

It is a non redox process and Ba(OH)2 has n factor of 2. It reacts with CO2 in molar ratio of 1:1 

∴ n factor of CO2 is 2

12. moles of Na2CO3 in 40 ml  of ‘s’ =

moles . of  NaOH in  40 ml is = = mole 

∴ In 60 c.c. of ‘S’ = moles of NaOH

Zn + 2 NaOH ⎯→ Na2ZnO2 + H2 

0.0105 moles of NaOH gives   moles of H2

∴Volume of H2 at S.T.P. = Lt = 0.1176 Lt   or 117 .6 ml

Mass of Na2CO3 in 100 ml = = 0.265

Mass of NaOH in 100 ml = 0. 7 g 

13. In this reaction, As2O3 acts as acidic oxide and NaHCO3 as a base, giving acid base neutralization reaction which is non – redox process. Here As2O3 does not act as basic oxide  in that case, it will form As2(CO3)3 which does not exist as non-metals do not form carbonates.

 

n-factor of As2O3 is 6 and that of NaHCO3 is 1. After the reaction As3+ is oxidised by I2 to As+5 while I2 is reduced to I–.

Normality of Na2S2O3.5H2O = = 0.1

Normality of I2 = Normality of Na2S2O3.5H2O =

∴ Equivalents of I2 = 0.1 ×22.4×10–3=Equivalent of  As3+ reacted in 25 ml=2.24 × 10–3

∴ Equivalents of As3+ reacted in 250 ml =  2.24 × 10–2

Moles of As3+ in 250 ml = = 1.12 × 10–2

Moles of As2O3 reacted = = 5.6 ×10–3

% Percentage of As2O3 = = 9.24%

14. Let the mol wt be  M

If each ZSM–5 has at least one Si atom 

= 28 ∴ M = 63.66 amu

If each ZSM–5 has at least one  Al atom 

= 27 ∴ M = 6058

∴ Minimum mol. wt of ZSM–5 = 6053. In which 1 Al atom and
= 95 Si atoms will be present. 

[Note: ZSM–5 has the formula the H[AlO2(SiO2)95]16H2O]

 

15. Equivalents of Barium permanganate = = 1.485 ×10–4

Equivalents of Fe2+ in excess = 1.485 × 10–4

Initial equivalents of Fe2+ = = 8 × 10–3

Equivalents of Fe2+ consumed by Cr6+ = 8 × 10–3 – 1.485 × 10–4 = 7.85 × 10–3

∴Equivalents of Cr6+ = 7.85 × 10–3

Equivalents of FeO. Cr2O3 = 7.85 × 10–3

Moles of FeO. Cr2O3 = = 1.3 × 10–3

Mass of FeO. Cr2O3 = 1.3 × 10–3 × 224 = 0.2912 gm.

% FeO.Cr2O3 = × 100 = 15. 57%

 

Objective Problems 

 

LEVEL – I

 

1. = ∴ Normality of H2SO4 solution 

=

Let V ml of water be added to 400 ml NH2SO4 to make it then,

=

V+400 = = 890.2 ml

V = 490.2 ml

∴ (D)

2. Equivalent of metal sulphate = Equivalent of metal

⇒

⇒ ⇒

⇒ ∴ EM = 12

∴ (C)

3. ROH (1 mol) + CH3MgBr ⎯→ CH4(1 mol) + Mg(OH)Br

As 1.12 ml of gas is obtained by 4.12 mg of alcohol

∴ 22400 ml of CH4 at STP = × 22400 mg × = 84.2 gm (Mol. Wt.)

∴ (C)

4. i) Molecule of O2 = = 0.5 × 10–23 gm
5. ii) One atom of N = = 2.3 × 10–23 gm

iii) 10–10 gm mol wt. of oxygen = 10–10 × 32 = 3.2 × 10–9 gm

6. iv) 10–10 gm atom of copper = 10–10 × 63 = 6.3 × 10–9 gm

∴ (A)

5. Ca3P2 + 6H2O ⎯→ 3Ca(OH)2 + 2PH3

∴ (C)

6. Normality of oxalic acid solution = = 0.4 N

N1V1 = N2V2 ⇒ 10 × 0.4 = V × 0.1

∴ V = 40 ml

∴ (A)

7. 3BaCl2 + 2Na3PO4 ⎯→ Ba3(PO4)2 + 6NaCl

Limiting reactant is Na3PO4

0.2 mol of Na3PO4 will give Ba3(PO4)2 = × 0.2 = 0.1 mol of Ba3(PO4)2

∴ (D)

8. Eq. of H2SO4 = 0.5 × 2 = 1.0

Eq. of Ca(OH)2 = 0.2 × 2 = 0.4

Equal equivalents reacts thus CaSO4 formed = 0.4

∴ Mole of CaSO4 formed = = 0.2

∴ (A)

9. Percent loss of H2O in one mole of Na2SO4xH2O

= = 55 ∴ x = 10

∴ (D)

10. 2KClO3 ⎯⎯ 2KCl + 3O2↑

245 gm of KClO3 on heating shows a weight loss of 96 gm (of O2)

∴ 100 gm of KClO3 on heating shows a weight loss = = 39.18%

∴ (C)

11. H3PO3 is dibasic acid, thus N = M × 2 = 2 × 0.3 = 6

∴ (D)

12. 5Fe2+ + MnO4– + 8H+⎯→ 5Fe3++ Mn2+ + 4H2O

5C2O42–+ 2MnO4– + 16H+ ⎯→ 10CO2 + 2Mn2+ + 8H2O

––––––––––––––––––––––––––––––––––––––––––––––––––––

5FeC2O4 + 3MnO4– + 24H+ ⎯→ 5Fe3+ + 10CO2 + 3Mn2+ + 12H2O

5 mol of FeC2O4 ≡ 3 mol of KMnO3 ∴ 1 mol of FeC2O4 =

∴ (A)

13.

∴ (A)

14. 40 × 0.246 × 8 = V × 0.154 × 3 (Meq. of S2O32– = Meq. of CrO42–)

∴ V = 170.4

∴ (C)

15.       Al2(SO4)        +  BaCl2 ⎯→ BaSO4↓ + AlCl3

Initial Meq.   20 × 0.2 × 6     20 × 0.6 × 2

              = 24                  = 24

Final Meq.    0                       0                 24            24

                    [Al3+]  = = 0.2 M

∴ (A)

16. 7 gms of sodium in 100 gm of acid dye

∴ 23 gms of sodium will be present in = = 328.5 (molecular of dye salt)

In acid dye one Na atoms replaces on H atom

∴ Molecular weight of acid dye = 328.5 – 23 + 1 = 306.5

∴ (D)

18. (B)
19. Ag2CO3 2Ag + CO2 + O2

Molecular weight of Ag2CO3 = 276 (2 × 108 + 60)

276 gm of Ag2CO3 gives Ag = 216 gm

∴ 2.76 gm Ag2CO3 gives Ag = 2.16 gm

∴ (A)

20. S22+⎯→ 2S6+ + 8e (n-factor =- 8)

Mn7+ + 3e +⎯→ MnO2 (n-factor = 3)

Meq. of KMnO4 = Meq. of S2O32–

⇒ 0.1 × 3 × V = × 1000 (molecular weight of Na2S2O3 = 158)

∴ V = 26.67 ml

∴ (B)

 

LEVEL – II

 

1. Meq, of NaBrO3 = 85.5 × 0.672 = 57.456

Let weight of NaBrO3 = W

∴ = 57.456 (equivalent weight = M/6) of n-factor = 6

∴ × 6 × 1000 = 57.456 ∴W = 1.45 gm

∴ (B)

2. Meq. of NaHSO3 = Meq. NaIO3= N × V = × 6 × 1000 (15+ + 62 → I–)

∴ × 2× 1000 = × 6 ×100

= 4.57

∴ (B)

3. O2–1 + 2e ⎯→ 2O–2 I2 + S2O32– ⎯→ 2I– + S4O62–

2I– ⎯→ I2 + 2e

Meq. of H2O2  = Meq. of I2 = Meq of Na2S2O3

N × 25 = 0.1 × 20, ∴ = 0.08

∴ (C)

4. KBr + NaBr          +  Ag+⎯→ AgBr

agm    (0.56 – a)                      0.97 gm

Applying POAc for Br atom 1 × nKBr  + 1 × nNaBr = 1 × nAgBr

∴ a = 0.1332 gm

∴ Fraction of KBr in the sample = = 0.2378

∴ (B)

5. Fe2S3: H2O : O2

Given mol ratio   1           2        3

Mole ratio in the reaction 2           6        3

Mole ratio in the reaction

That is limiting reagent is H2O

Moles of Fe(OH)3 produced by 2 moles of H2O = × 2 = 1.34

∴ (C)

6. Let amount of Na2CO3 in 1 gm of mixture be a and since Na2CO3 and NaHCO3 are in equimolar amounts.

∴ a = 0.588 gm

∴ Weight of NaHCO3 = (1 – 0.558) = 0.442 gm

Met of Na2CO3 + Meq. of NaHCO3 = Meq. of HCl

∴ = 0.1 × V

∴ V = 157.9 ml

∴ (A)

7. I2 + 2Cl2 ⎯→ ICl + ICl3

25.4 gm of Iodine contains 0.1 moles of it

14.2 gm of chlorine contains 0.2 moles of it

∴ Moles of ICl & ICl3 produced will be 0.1 and 0.1. Hence Molar ratio 1: 1

∴ (A)

8. 2Fe + 3H2O ⎯→ Fe2O3

3 × 18gm of steam will react with 2 × 56 gm of Iron

∴ 18 gm of steam will convert = = 37.3 gm of Fe

∴ (A)

9 C12H22O11 + 12O2 ⎯→ 12CO2 + 11H2O 

1 mole of starch requires 12 mole of oxygen 

∴ mole of starch requires = 916.2 gm 

∴ (A)

10 2H2O2 ⎯→ H2O + O2

22400 ml of O2 evolved from 68 gm of H2O2 

∴ 10 ml of O2 is evolved from gm of H2O2 

Hence 1 ml  of H2O2 contain mol = 0.00178 equivalent 10 ml of H2O2 will 0.0178 equivalents which will be  present in 100 ml of KMnO4 solution.

Amount of KMnO4 in given sample = = 0.563 gm

∴ (B)

11. Let us  first solve this problem by writing the complete balanced reaction.

3BaCl2 + 2 Na3PO4 ⎯→ Ba3(PO4)2 ↓+ 6NaCl

We can see that the moles of BaCl2 used is times the moles of Na3PO4.
∴ to react  with 0.2 mol of Na3PO4, the moles of BaCl2 required would be 
0.2 × = 0.3. Since BaCl2 is 0.5 mol, we can conclude  that Na3PO4 is the limiting reagent. Therefore, moles of Ba3(PO4)2 formed is 0.2 × = 0.1 mol.

∴ (D) 

 

12. The above reaction can be balanced by using the ion electron method as under:

Oxidation reaction : ⎯→ CO2

Reduction reaction : ⎯→

Balancing atoms other than O 

⎯→ 2 CO2

⎯→  Mn2+

Since medium is acidic 

⎯→ 2CO2  (oxygen is already balanced)

8H+ + ⎯→ Mn2+ + 4H2O

Balancing Charge 

⎯→ 2CO2 + 2e– …(1)

5e– + 8H+ + ⎯→ Mn2+ + 4H2O …(2)

Multiplying equation (1) by 5 and equation (2) by 2, and adding, we get 

+ + 16H+ ⎯→ 10CO2 + 2Mn2+ + 8H2O

∴ (A) 

13. ‘n’ factor of ClO– is 2 ( Cl+1 → Cl–1)

∴ Equivalents of ClO– =  0.15 × 2 = 0.3

If the molecular weight of the chromium oxide is M, and the oxidation state of chromium in it is x. Then equivalents of chromium oxide is 

(6–x) × ( number of chromium atoms in the oxide is 1 Q all the choices have 1 Cr per molecule)

∴ (6–x) × = 0.3

= or  =

Put  each choice and find out the correct result. Choice (a) and (C) can be ignored ( because oxidation state of Cr is +6 and +8 respectively. The later is an impossible oxidation state  of Cr).

∴ (B)

14. Moles of H2XO4 = [ Mx = Atomic mass of X]

‘n’ factor of H2XO4 = 2 [QH2XO4 is a dibasic acid] ∴ 2.56 × 10–3 = × 2

Mx = 31.96 g / mol ≈ 32 g / mol

∴ X is Sulphur 

Sulphur has 16 protons and 16 neutrons (for these type of problems, atomic masses and atomic numbers would be supplied)

∴ (A)

15. 8 g of S = mole = mole 

∴ Moles of BaSO4 is  also , as the mass of S is  conserved. 

∴ (D)

16 Moles of  CaO =

⇒ Moles of CaCl2 =

⇒ Mass of CaCl2 = 111 = 3.21 gm

⇒ % = 100 = 32.1%

∴ (B)

17. The ‘n’ factor of KMnO4 is 5 while that of K2Cr2O7 is 6. So for the same number of moles, K2Cr2O7 will have greater equivalence than KMnO4.

∴ (B)

18. ‘n’ factor of is 6.

Equivalents of needed = equivalents of N2H5+ = 0.136

∴ Mole of   = = 0.0227

∴ (D)

 

19. Percentage of  oleum = Mass of total H2SO4 produced by 100 gm of oleum.

Increase in mass of oleum is due to 

H2O. (SO3 + H2O ⎯→ H2SO4) which  is  9 gm.

Moles of H2O = =

Moles of SO3 =

Mass of SO3 = × 80 = 40

∴  (A)

20. n–factor of FeC2O4 is 3 as

Fe2+ ⎯→ Fe3+ and C2O4– – ⎯→ CO2

n = 1        n = 2

∴ =

V = = 1200 ml = 1.2 Litre 

∴ (B)

 

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