(a) Writing the On on each atom above its symbol, then
\(3\overset { +1 }{ H } \overset { -1 }{ Cl } (aq)+\overset { +1 }{ H } \overset { +5 }{ N } \overset { -2 }{ { O }_{ 3 } } (aq)\longrightarrow \overset { 0 }{ { Cl }_{ 2 }(g)+ } \overset { +3 }{ N } \overset { -2 }{ O } \overset { -1 }{ Cl(g)+ } \overset { +1 }{ { 2H }_{ 2 } } \overset { -2 }{ O } (l)\)
Here, the On of Cl increases from -1 in HCl to O in Cl₂ , therefore, Cl^- is oxidised and hence, HCl acts as the reducing agent. The ON of N decreases from +5 in HNO₃ to +3 in NOCL, therefore, HNO₃ acts as the oxidising agent. Thus this reaction is a redox reaction.
(b) Writing the ON of each atom above its symbol, we have,
\(\overset { +2 }{ Hg } \overset { -1 }{ { Cl }_{ 2 } } (aq)+\overset { +1 }{ 2K } \overset { -1 }{ I } (aq)\longrightarrow \overset { +2 }{ Hg } { \overset { -1 }{ I } }_{ 2 }(s)+2\overset { +1 }{ K } \overset { -1 }{ { Cl }^{ - }(aq) } \)
Here, the On of none of the atoms undergo a change, therefore, this reaction is not a redox reaction.
(c) \(\overset { +3 }{ { Fe }_{ 2 } } { \overset { -2 }{ O } }_{ 3 }(s)+3\overset { +2 }{ C } \overset { -2 }{ O } (g)\ \overset { \Delta }{ \longrightarrow } \ 2\overset { 0 }{ Fe } (s)+3\overset { +4 }{ C } \overset { -2 }{ { O }_{ 2 } } (g)\)
Here, On of decreases from +3 in Fe₂O₃ to 0 in Fe, therefore, Fe₂O₃ acts as an oxidising agent. Further, On of C increases from +2 in CO to +4 in CO₂, therefore, CO acts as a reducing agent. Thus, this reaction is an example of redox reaction.
(d) Writing the ON of each atom above its symbol, then
\(\overset { +3 }{ P } { \overset { -1 }{ Cl } }_{ 3 }(l)+3\overset { +1 }{ { H }_{ 2 } } \overset { -2 }{ O } (l)\quad \longrightarrow \quad 3\overset { +1 }{ H } \overset { -1 }{ Cl } (aq)+\overset { +1 }{ { H }_{ 3 } } \overset { +3 }{ P } \overset { -2 }{ { O }_{ 3 } } (aq)\)
Here, On of none of the atoms undergo a change, therefore, this reaction is not a redox reaction.
(e) Writing the ON of each atom above its symbol, then
\(4\overset { -3 }{ N } \overset { +1 }{ { H }_{ 3 } } (aq)+3\overset { 0 }{ { O }_{ 2 } } (g)\quad \longrightarrow \quad 2\overset { 0 }{ { N }_{ 2 } } (g)+6\overset { +1 }{ { H }_{ 2 } } \overset { -2 }{ O } (l)\)

The balanced equation for the reaction is
\(\underset { \overset { 4\quad \times \quad 17 }{ =68g } }{ 4{ NH }_{ 3 }(g) } +\underset { \overset { 5\quad \times \quad 32 }{ =160g } }{ { 5O }_{ 2 }(g) } \longrightarrow \underset { \overset { 4\quad \times \quad 30 }{ =120g } }{ { 4NO(g) } } +{ 6H }_{ 2 }O(g)\)
Here, 68 g of NH₃ will react will 02 = 160 g
\(\therefore\) 10 g of NH3 will react with 0₂  = \(\frac { 160g }{ 68g } \) x 10g = 23.6g
But the amount of 02 which is actually available is 20.0 g which is less than the amount which is needed. Therefore, 02 is the limiting reagent and hence calculations must be based upon the amount of 02 taken and not on the amount of NH3 taken. From the equation
160 g of O₂ produce NO = 120 g
\(\therefore\) 20g of O₂ will produce NO = \(\frac { 120 }{ 160 } \times 20\) = 15g