The ratio of coefficient of HNO3- Fe NO 3 2 and NH 4 NO 3 in the following redox equation- Fe- HNO 3 - Fe NO 3 2 - NH 4 NO 3 - H 2 O in the balanced form will be-

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Ion Electron Method

The ratio of ...

Question

The ratio of coefficient of ${H N O}_{3}, {F e ({N O}_{3})}_{2}$ and ${N H}_{4} {N O}_{3}$ in the following 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 form will be?

10 : 1 : 4

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10 : 4 : 1

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4 : 10 : 1

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4 : 1 : 10

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

H N O_{3}, F e (N O_{3})_{2}, N H_{4} N O_{3}

N O_{3}^{⊖} + Z n ⟶ {Z n}^{2 +} + N H_{4}^{\oplus}

N O_{3}^{⊖} + 7 H_{2} O + 4 Z n ⟶ N H_{4}^{\oplus} + 10 ⊖ O H + 4 {Z n}^{2 +}

Balancing.1-HNO3+Can(oh)2->can(No3)2+h2o

H N O_{3}

\Rightarrow

C u (N O_{3})_{2}

H_{2} O

H N O_{3}

\Rightarrow

Z n (N O_{3})_{2}

H_{2} O

I_{2}

H N O_{3}

\Rightarrow

N O_{2}

H_{2} O

I : Z n + d i l . H_{2} S O_{4} ⟶ Z n S O_{4} + H_{2}

I I : Z n + c o n c . H_{2} S O_{4} ⟶ Z n S O_{4} + S O_{2} + H_{2} O

I I I : Z n + H N O_{3} ⟶ Z n {({N O}_{3})}_{2} + {N H}_{4} {N O}_{3} + H_{2} O

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