Which of the following correctly represents the equivalent mass in a redox change of reaction, ${(N H_{4})}_{2} C r_{2} O_{7} ⟶ N_{2} + C r_{2} O_{3} + 2 H_{2} O$ ?

E_{{(N H_{4})}_{2} C r_{2} O_{7}} = \frac{M}{6}

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E_{N_{2}} = \frac{M}{6}

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E_{C r_{2} O_{3}} = \frac{M}{6}

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E_{N_{2}} = \frac{M}{3}

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Solution

The correct options are
A $E_{{(N H_{4})}_{2} C r_{2} O_{7}} = \frac{M}{6}$
B $E_{N_{2}} = \frac{M}{6}$
D $E_{C r_{2} O_{3}} = \frac{M}{6}$
${(N H_{4})}_{2} C r_{2} O_{7} ⟶ N_{2} + C r_{2} O_{3} + 2 H_{2} O$
-3 +6 0 +3
As we know,
Equivalent weight $= \frac{M o l a r w e i g h t}{n - f a c t o r}$
$E_{{(N H_{4})}_{2} C r_{2} O_{7}} = \frac{M}{6}$
$E_{N_{2}} = \frac{M}{6}$
$E_{C r_{2} O_{3}} = \frac{M}{6}$

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Similar questions

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

E_{C r_{2} O_{3}} = \frac{M}{x}

x

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

B r_{2}

3 B r_{2} + 6 O H^{-} \to 5 B r^{-} + B r O_{3}^{-} + 3 H_{2} O

H_{3} P O_{3}

H_{3} P O_{3} \to H_{3} P O_{4} + P H_{3}

H_{3} B O_{3}

H_{3} B O_{3})

N a_{2} B_{4} O_{7}

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