The ratio of coefficients of $H N O_{3}, F e {(N O_{3})}_{2}$ and $N H_{4} N O_{3}$ in the redox equation $F e + H N O_{3} \to F e {(N O_{3})}_{2} + N H_{4} N O_{3} + H_{2} O$ in the balanced from will be

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10 : 1 : 4

4 : 1 : 10

4 : 10 : 1

10 : 4 : 1

answer is B.

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Detailed Solution

$\overset{0}{F e} + \overset{+ 5}{H N O_{3}} \to \overset{+ 2}{F e} {(N O_{3})}_{2} + \overset{- 3}{N H_{4}} N O_{2} + H_{2} O$

$(i) \overset{0}{F e} \to \overset{+ 2}{F e} {(N O_{3})}_{2} ↓ 2$

$(i i) \overset{+ 5}{H N O_{3}} \to \overset{- 3}{N H_{4}} N O_{3} ↓ 8$

To make increase in O.N.=decrease, multiply eqn.

(i) by 4 and add. Hence

$4 F e + H N O_{3} \to 4 F e {(N O_{3})}_{2} + N H_{4} N O_{3}$

To balance N atoms on the two sides, multiply $H N O_{3}$ by 10. Hence

$4 F e + 10 H N O_{3} \to 4 F e {(N O_{3})}_{2} + N H_{4} N O_{3}$

To balance O-atoms, add $3 H_{2} O$ on RHS. Hence

$4 F e + 10 H N O_{3} \to$

$4 F e {(N O_{3})}_{2} + N H_{4} N O_{3} + 3 H_{2} O$

H atoms are automatically balanced.

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