What will be the n-factor of the reactant in the following reaction ?
$(N H_{4})_{2} C r_{2} O_{7} Δ - \to N_{2} + C r_{2} O_{3} + 4 H_{2} O$

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Solution

The oxidation number of Cr in ammonium dichromate is +6.
The oxidation number of Cr in chromium trioxide is +3.
The change in the oxidation number of one Cr atom is 3.
The overall change in the oxidation number for 2 Cr atoms is 6.
Hence, the n factor is 6.

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Similar questions

Q. The reaction

(N H_{4})_{2} C r_{2} O_{7} \to N_{2} + C r_{2} O_{3} + 4 H_{2} O

, is an example of :

Q. The following intramolecular redox reaction shows a loss and gain of how much electrons?

(N H_{4})_{2} C r_{2} O_{7} \to N_{2} + C r_{2} O_{3} + 4 H_{2} O

(NH₄)₂Cr₂O₇ $\overset{}{\to}$ N₂ + Cr₂O₃ + 4H₂O. What volume of nitrogen at STP, will be evolved when 63 g of ammonium dichromate is decomposed? (H = 1, N = 14, O = 16, Cr = 52)

Q. The elements undergoing reduction and oxidation during the reaction,

(N H_{4})_{2} C r_{2} O_{7} \to N_{2} + C r_{2} O_{3} + 4 H_{2} O

are respectively:

(N H_{4})_{2} C r_{2} O_{7} \to C r_{2} O_{3} + 4 H_{2} O + N_{2}

The mass of

C r_{2} O_{3}

formed by the decomposition of 63 gm of ammonium dichromate :

[N = 14, H = 1, O = 16, C r = 52]

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What will be the n-factor of the reactant in the following reaction ?(NH4)2Cr2O7Δ−→N2+Cr2O3+4H2O

The oxidation number of Cr in ammonium dichromate is +6.The oxidation number of Cr in chromium trioxide is +3.The change in the oxidation number of one Cr atom is 3.The overall change in the oxidation number for 2 Cr atoms is 6.Hence, the n factor is 6.

What will be the n-factor of the reactant in the following reaction ?
$(N H_{4})_{2} C r_{2} O_{7} Δ - \to N_{2} + C r_{2} O_{3} + 4 H_{2} O$

The oxidation number of Cr in ammonium dichromate is +6.
The oxidation number of Cr in chromium trioxide is +3.
The change in the oxidation number of one Cr atom is 3.
The overall change in the oxidation number for 2 Cr atoms is 6.
Hence, the n factor is 6.