Given : \(X\,\mathrm{Na_2HAsO_3} + Y\,\mathrm{NaBrO_3} + Z\,\mathrm{HCl} \rightarrow \mathrm{NaBr} + \mathrm{H_3AsO_4} + \mathrm{NaCl}\)
The values of X, Y and Z in the above redox reaction are respectively :
The values of X, Y and Z in the above redox reaction are respectively :
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- 2, 1, 2
- 2, 1, 3
- 3, 1, 6
- 3, 1, 4
The Correct Option is C
Solution and Explanation
Equalise electrons: \(2X = 6Y\Rightarrow X : Y = 3 : 1\), i.e. X = 3, Y = 1.
Na balance: left \(2X + Y = 7\); right NaBr (1) + NaCl (n) gives n = 6 NaCl.
Cl balance: Z = 6.
Check H: left \(3 + 6 = 9\); right \(3\times3 = 9\). O: left \(9+3 = 12\); right \(12\). Balanced.
\(3\,\mathrm{Na_2HAsO_3} + \mathrm{NaBrO_3} + 6\,\mathrm{HCl}\rightarrow \mathrm{NaBr} + 3\,\mathrm{H_3AsO_4} + 6\,\mathrm{NaCl}\).
Answer: (3) 3, 1, 6
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