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Standard XII

Chemistry

Normality

25.0 mL of 0....

Question

25.0 mL of 0.05 M solution of $N H_{2} O H$ was boiled with excess of $F e^{3 +}$ in acidic medium. The $F e^{2 +}$ formed required 25.0 mL of 0.1 N $C r_{2} O_{7}^{2 -}$ in acidic medium. What is oxidation number of the $N$ in new product?

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Solution

1 equivalent of $F e^{3 +}$ will react with 1 equivalent of $K_{2} C r_{2} O_{7}$ .
The number of milliequivalents of $K_{2} C r_{2} O_{7} = 0.1 N \times 25.0 m L = 2.5 m e q$ .
This corresponds to the number of milliequivalents of $F e^{3 +}$ .
The number of moles of $N H_{2} O H$ is $0.05 N \times 25.0 m L = 1.25 m e q$ .
Thus, 2 moles of $F e^{3 +}$ will combine with 1 mole of $N H_{2} O H$ .
Thus, the change in the oxidation number will be 2 i.e, from -1 to +1.
$4 F e^{3 +} (a q) + 2 N H_{2} O H (a q) \to 4 F e^{2 +} (a q) + N_{2} O (g) + H_{2} O (l) + 4 H^{+} (a q)$

The oxidation number in $N$ in new product is $+ 1$ .

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25.0 mL of 0.05 M solution of NH2OH was boiled with excess of Fe3+ in acidic medium. The Fe2+ formed required 25.0 mL of 0.1 N Cr2O2−7 in acidic medium. What is oxidation number of the N in new product?

25.0 mL of 0.05 M solution of $N H_{2} O H$ was boiled with excess of $F e^{3 +}$ in acidic medium. The $F e^{2 +}$ formed required 25.0 mL of 0.1 N $C r_{2} O_{7}^{2 -}$ in acidic medium. What is oxidation number of the $N$ in new product?