(i) The entire quantity of ammonia is consumed in the reaction. So ammonia is the Iimiting reagent. Some quantity of CO₂ remains unreacted, so CO₂ is the excess reagent.
(ii) Quantity of urea formed = number of moles of urea formed x molar mass of urea
= 19 moles x 60 g mol^-1
= 1140 g = 1.14 kg
Excess reagent leftover at the end of the reaction is carbon dioxide.
Amount of carbon dioxide leftover
= number of moles of CO₂ left over x molar mass of CO₂   
= 7 moles x 44 g mol^-1 
= 308 g

The skeletal equation is:
\(Cl_{ 2{ (aq) } }+SO_{ 2(aq) }+H_{ 2 }O_{ (1) }\rightarrow { Cl }^{ - }_{ (aq) }+\)\({{SO}^{2}_{4}}_{(aq)}\)
Reduction half equation:
\(Cl_{2(aq)}\rightarrow {Cl}^-_{(aq)}\)
Balance CI atoms,
\(\overset{0}Cl_{2(aq)}\rightarrow \overset{-1}2Cl^{-1}_{(aq)}\)
Balance O.N. by adding electrons
\(Cl_{2(aq)}+2e^-\rightarrow 2Cl^-_{(aq)}\)
Oxidation half equation:
\(\overset{+4}{{{SO}_{{2}_{(aq)}}}}\rightarrow{SO}_{{4}^{2-}_{(aq)}}+2e^{-}\)
Balance O.N. by adding electrons
SO_{2 (aq)} \(\rightarrow\)SO₄^2-_(aq) + 2e^-
Balance charge by adding 4H⁺ ions:
SO_{2 (aq)} \(\rightarrow\)SO₄^2-_(aq) + 4H⁺ _(aq) + 2e^-
Balance O atoms by adding 2H₂O,
SO_{2 (aq)} + 2H₂O_(l) \(\rightarrow\)SO₄^2-_(aq) + 4H⁺ _(aq) + 2e^-
Adding equation (1) and (2), we have,
CI_{2 (aq)}+ SO_{2 (aq)}+ 2H₂O_(I) \(\rightarrow\)2CI^-_(aq) + SO₄^2-_(aq) + 4H⁺ _(aq)
This represents the balanced redox reaction.

The skeletal equation is:
\(Mn^{3+}_{aq} \rightarrow Mn_{aq}^{2+}+MnO_{2(s)}+H^+_{(aq)}\)
Oxidation half equation:
\(\overset{+3}Mn^{3+}_{(aq)}\rightarrow \overset{+4}MnO_2{s}\)
Balance O.N. by adding electrons,
\(Mn^{3+}_{(aq)}\rightarrow Mn{O}_{2{(s)}}+e^-\)
Balance charge by adding 4H⁺ ions,                   
\(\overset{3+}Mn_{(aq)}\rightarrow MnO_{2{(s)}}+4H^+_{(aq)}+e^-\)
Balance O atoms by adding 2H₂O
\(\overset{ 3+}Mn_{(aq)}+2H_2O_{(1)}\rightarrow MnO_{2{(s)}}+4H^+_{(aq)}+e^-\)....(1)
Reduction half equation:
\(\overset{3+}Mn^{3+}\rightarrow\overset{2+} Mn^{2+}\)
Balance O.N. by adding electrons:
\(Mn^{3+}_{(aq)}+e^-\rightarrow Mn^{2+}_{(aq)}\).....(2)
Adding Equation (1) and (2), the balanced equation for the disproportionation reaction is
\(2M{n}^{3+}_{(aq)}+2{H}_{2}{O}_{(1)}\rightarrow Mn{O}_{2(s)}+M{n}^{2+}_{(aq)}+{4H}^{+}_{(aq)}\)

(a) (i) CI oxidation number 0 in Cl₂.
(ii) CI oxidation number -1 in HCI.
(iii) Cl oxidation number +3 in HCIO₂.
(iv) Cl oxidation number +5 in KClO₃.
(v) CI oxidation number +7 in Cl₂O₇.
(b) In H₂O₂, oxidation number of oxygen is -1 and it can vary from 0 to -2 (+2 is possible in OF₂).
The oxidation number can decrease or increase, because of this H₂O₂ can act both oxidising and reducing agent.
Ozone (O₃) only acts as oxidising agent since it decomposes to give nascent oxygen.

Flask A:
1 mole of oxygen gas = 6.023 x 10²³ molecules
\(\therefore\) 0.5 mole of oxygen gas = 6.023 x 10²³ x 0.5 molecules
The number of atoms in flask A = 6.023 x 10²³ x 0.5 x 2
= 6.023 x 10²³ atoms.
Flask B:
1 mole of ozone gas = 6.023 x 10²³ molecules
\(\therefore\) 0.4 mole of ozone gas = 6.023 x 10²³ x 0.4 molecules
The number of atoms in flask B = 6.0²³ x 1023 x 0.4 x 3
= 7.227 x 10²³atoms.
\(\therefore\) Flask B contains a great number of oxygen atoms as compared to flask A.

Net ionic reaction:
\(Mn{O}_4^-+Fe^{2+}+H^+\rightarrow Mn^{2+}+Fe^{3+}\)
Oxidation half reaction:
\(Fe^{2+}\rightarrow Fe^{3+}+e^-\) ....(1) 
Reduction half reaction:
\(Mn{O}_4^-+5e^-\rightarrow Mn^{2+}\)...(2)

To balance O and H, H⁺ and H₂O are added.
\(Mn{O}_4^-+5Fe^{2+}+8H^{+}\rightarrow Mn^{2+}+5Fe^{3+}+4H_2O\)

(ii) Balance the changes in O, N, by multiplying the oxidant and reductant by suitable numbers.
4MnO₂ + 2O₂ + KOH\(\rightarrow\)K₂MnO₄ + H₂O
(iii) Balance the equation atomically (except O and H).
4MnO₂ + 2O₂ + KOH\(\rightarrow\) 4K₂MnO₄ + H₂O
(iv) Balance oxygen and hydrogen atoms by multiplying H₂O by 4.
4MnO₂ + 2O₂ + 8KOH \(\rightarrow\) 4K₂MnO₄ + 4H₂O

(ii) 2KMnO₄ +10 FeSO₄ + H₂SO₄ \(\rightarrow\)K₂SO₄ + MnSO₄ + Fe₂(SO₄)₃ + H₂O
(iii) Balance the equation atomically (except O and H) and sulphate ions.
2KMnO₄ + 10 FeSO₄ + 8 H₂SO₄ \(\rightarrow\)K₂SO₄ + 2MnSO₄ + 5 Fe₂(SO₄)₃ + H₂O
(iv) BalanceO atoms by multiplying H₂O by 8.
2KMnO₄ + 10 FeSO₄ + 8 H₂SO₄ \(\rightarrow\)K₂SO₄ + 2MnSO₄ + 5 Fe₂(SO₄)₃ + 8 H₂O

(a) Equivalent mass of an acid:
Equivalent mass of an acid is the number of parts by mass of the acid which contains 1.008 part by mass of replaceable hydrogen atom.
Equivalent mass of an acid = \(\frac { Molar\ mass\ of\ the\ acid }{ Basicity\ of\ the\ acid } \)
(b) Equivalent mass of a base:
It is defined as the number of parts by mass of the base which contains one replaceable hydroxyl ion or which completely neutralizes one gram equivalent of an acid.
Equivalent mass of a base = \(\frac { Molar\ mass\ of\ the\ base }{ Acidity\ of\ the\ base } \)
(c) Equivalent mass of an oxidising agent :
It is defined as the number of parts by mass of an oxidising agent which can furnish 8 parts by mass of oxygen for oxidation.
(d) Equivalent mass of a reducing agent:
It is defined as the number of parts by mass of the reducing agent which is completely oxidised by 8 parts by mass of oxygen or with one equivalent of any oxidising agent.

Ion-electron method makes use of the Half reactions. Steps involved in this method are,
1. Write the equation in the net ionic form without attempting to balance it.
2. Write and locate the oxidation number of atoms undergoing oxidation and reduction from the knowledge of calculation of oxidation number.
3. Write two half reactions showing oxidation and reduction separately.
4. Balance oxygen atoms by adding required number of water molecules to the side deficient in oxygen atoms.
5. Add required number of H⁺ ions to the side deficient in hydrogen atom if the reaction is in acidic medium
6. Add electrons to whichever side is necessary to make up the difference in oxidation number.
7. Add the two half reactions. The resulting equation is a net balanced equation.
8. For reactions in basic medium, add H₂O and hydrogen ion to balance H and O.
9. Finally, balance the equation by cancelling common species present on both sides of the equation.

(ii) 2KMnO₄ + 10 HCI \(\rightarrow\) KCl + MnCl₂ + H₂O + Cl₂
(iii) Balance the equation atomically (except O and H).
2KMnO₄ + 10 HCI \(\rightarrow\) 2KCl + 2MnCl₂ + H₂O + Cl₂
(iv) Balance chlorine atoms by adding HCI and multiplying Cl₂ by 5.
2KMnO₄ + 16HCI \(\rightarrow\) 2KCl + 2MnCl₂ + H₂O + 5Cl₂
(v) To balance O and H, H₂O is multiplied by 8.
2KMnO₄ + 16HCI \(\rightarrow\) 2KCl + 2MnCl₂ + 8 H₂O + 5Cl₂

(ii) Balance the changes in O.N. by multiplying the oxidant and reductant by suitable numbers
2 C₆H₆ + 15 O₂\(\rightarrow\)CO₂ + H₂O
(iii) Balance the equation atomically (except O and H).
2 C₆H₆ + 15 O₂\(\rightarrow\)12 CO₂ + H₂O
(iv) Balance O atoms by adding one H₂O molecule to the RHS for making the number of molecules of H₂O to be 6.
2 C₆H₆ + 15 O₂\(\rightarrow\)12 CO₂ + 6 H₂O

(a) Oxidation number refers to the number of charges an atom would have in a molecule or an ionic compound, if electrons were transferred completely.
(b) Rules to assign oxidation number:
1. Oxidation number of a substance in its elementary state is equal to zero (H₂, Br₂, Na)
2. Oxidation number of a mono-atomic ion is equal to the charge on the ion (Na⁺= + 1,S^2- = -2).
3. Oxidation number of hydrogen in a compound is + 1 (except hydrides).
4. Oxidation number of hydrogen in metal hydrides is -1 (NaH, CaH₂).
5. Oxidation number of oxygen in a compound is -2 (except OF₂ and peroxides).
6. Oxidation number of oxygen in peroxides is -1 (H₂O₂, Na₂O₂) = - 1.
7. Oxidation number of oxygen in fluorinated compounds is either + 1 or +2.
(OF₂ = +2, O₂F₂ = + 1).
8. Fluorine has an oxidation number -1 in all its compounds.
9. The sum of the oxidation number of all the atoms in neutral molecules is equal to Zero.
10. For all ions, the sum of the oxidation number of all atoms is equal to the charge of the ion.

(a)2NH_3(g) + CO_2(g) \(\rightarrow\) (NH₄)₂CO_(ag) + H₂O_(I)
2 moles             1 mole               1 mole
No. of moles of ammonia =\({637.2\over 17}=37.45\) mole
No. of moles of CO₂ \(={1142\over 44}=25.95\) moles
As per the balanced equation, one mole of CO₂ requires 2 moles of ammonia.
\(\therefore\) No. of moles of NH₃ required to react with 25.95 moles of CO₂ is
\(={2\over 1}\times 25.95=51.90\) moles
\(\therefore\) 37.45 moles of NH₃ is not enough to completely react with CO₂ (25.95 moles).
Hence, NH₃ must be the limiting reagent, and CO₂ is excess reagent.
(b) 2 moles of ammonia produce 1 mole of urea.
\(\therefore\) Limiting reagent 37.45 moles of NH₃ can produce \({1\over 2}\times 37.45\)moles of urea.
= 18.725 moles of urea.
\(\therefore\) The mass of 18.725 moles of urea = No. of moles x Molar mass
= 18.725 x 60
= 1123.5g of urea.
(c) 2 moles of ammonia requires 1 mole of CO₂,
\(\therefore\)Limiting reagent 37.45 moles of NH₃ will require \({1\over 2}\times 37.45\) moles of CO₂.
= 18.725 moles of CO₂
\(\therefore\) No. of moles of the excess reagent (CO₂) left = 25.95 - 18.725 = 7.225
The mass of the excess reagent (CO₂) left = 7.225 x 44
= 317.9g of CO₂·

Molar mass of Mg CO₃ = 84.32 g mol^-1
Percentage of Mg =\({24\over 84.32}\times 100=28.46\%\)
Percentage of C=\({12\over 84.32}\times 100=14.23\%\)
Percentage of O₃=\({48\over 84.32}\times 100=57.0\%\)
\(\underset{84.32\ g}{MgCO_3}\rightarrow MgO+\underset{44\ g}{CO_2 }\)
84.32 g of 100% pure MgCO₃ gives 44g of CO₂.
\(\therefore\) 100 x 10³ g of 100 % pure MgCO₃ gives =\({44\over 84.32}\times 100\times 10^3\)
= 52.182 x 10³g CO₂ 
100% pure MgCO₃ gives 52.182 x 10³ g CO₂
\(\therefore\) 90% pure MgCO₃ will give \({52.182\times 10^3\over 100}\times 90\) = 46963.8 g CO₂
= 46.96 Kg CO₂.

\(\therefore\) The empirical formula is C₁₄H₉Cl₅.
Calculation of Molecular formula:
The empirical formula mass(C₁₄H₉Cl₅) = \((14\times 12)+(9\times1)+(5\times 35.5)\)
= 168 + 9 + 177.5 = 354.5
\(n=\frac{Molecular\ mass}{Empirical\ formula\ mass}=\frac{354.5}{354.5}=1\)
Molecular formula = (Empirical formula)_n
= (C₁₄H₉Cl₅)_l
\(\therefore\) Molecular formula = C₁₄H₉Cl₅

\(\therefore\) The empirical formula is C₉H₈O₄.

\(\therefore\) The empirical formula = CH₂Cl

(a) The equivalent mass of a reducing agent is the number of parts by mass of the reducing agent which is completely oxidised by 8 parts by mass of oxygen or one equivalent of any oxidising agent.
(b) Ferrous sulphate is a reducing agent
\(\underset{2\times 152g}{2FeSO_4}+H_2SO_4+\underset{16g}{[O]}\rightarrow Fe_2{(SO_4)}_3+H_2O\)
16 parts by mass of oxygen oxidised 304 parts by mass of FeSO₄.
\(\therefore\) 8 parts by mass of oxygen will oxidise \(\frac{316}{80}\times8\) parts by mass of ferrous sulphate = 152
The equivalent mass of ferrous sulphate (anhydrous) = 152.
The equivalent mass of crystalline ferrous sulphate is (FeSO₄.7H₂O) = 152 + 126 = 278
The equivalent mass of crystalline ferrous sulphate = 278.

(a) The equivalent mass of an oxidizing agent is the number of parts by mass which can furnish 8 parts by mass of oxygen for oxidation.
(b) Potassium permanganate is an oxidizing agent 
\(\underset { \overset { 2\times 158 }{ 316\ g } }{ 2\ KMn{ O }_{ 4 } } +3H_2SO_4\rightarrow K_2SO_4+2MnSO_4+3H_2O+\underset { \overset { 5\times 16 }{ 80g } }{ 5\left[ O \right] } \)
\(\therefore\) 8 parts by mass of oxygen will be furnished by \(\frac{316}{80}\times=31.6\)
Equivalent mass of KMnO₄ = 31.6g eq^-1.

\(Cr{(OH)}_4^-+H_2O_2\rightarrow Cr{O}_4^{2-}\)
Oxidation half reaction:
\(\underset{(+3)}{Cr{(OH)}_4^-}\rightarrow\underset{(+6)} {Cr{O}_4^{2-}}+3e^-\).... (1)
Reduction half reaction:
\(H_2O_2+e^-\rightarrow H_2O\)....(2)
(+2)                        (+1)

To balance oxygen and hydrogen atoms, OH^- and H₂O are added.
\(2Cr{(OH)}_4^-+3H_2O_2+2OH^-\rightarrow 2Cr{O}_4^{2-}+8H_2O\)

\(Mn{O}_4^-+Fe^{2+}\rightarrow Mn^{2+}+Fe^{3+}\)
Oxidation half reaction:
\(Fe^{2+}\rightarrow Fe^{3+}+e^-\).....(1)
Reduction half reaction:
\(Mn{O}_4^-+5e^-\rightarrow Mn^{2+}\).......(2)
(+7)

To balance oxygen, H⁺ is added on RHS and H₂O is added on LHS.
\(Mn{O}_4^-+5Fe^{2+}+8H^+\rightarrow Mn^{2+}+5Fe^{3+}+4H_2O\)

\(Mn{O}_4^-+I^-\rightarrow MnO_2+I_2\)
Oxidation half reaction:
\(2I^-+2e^-\rightarrow I_2\).......(1)
Reduction half reaction:
\(Mn{O}_4^-\rightarrow MnO^2+3e^-\).......(2)
Equation (1) is multiplied by 3

To balance oxygen and hydrogen atoms, H⁺ is added on RHS and H₂O is added on LHS.
\(2Mn{O}_4^-+6I^-+4H^+\rightarrow 2MnO_2+2I_2+2H_2O\)
In acidic medium