In the reaction:
$C r_{2} O_{3} + N a_{2} C O_{3} + K N O_{3} ⟶ N a_{2} C r O_{4} + C O_{2} + K N O_{2}$
sum of equivalent weights of $K N O_{2}$ and $N a_{2} C r O_{4}$ produced is .

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Solution

Molecular weight of $K N O_{2} = 85 g / m o l$
Molecular weight of $N a_{2} C r O_{4} = 162 g / m o l$
$C r_{2} O_{3} + N a_{2} C O_{3} + K N O_{3} ⟶ N a_{2} C r O_{4} + C O_{2} + K N O_{2}$
Here species which are undergoing change in oxidation number are $C r$ and $N$ .
$Oxidation half : {+ 3 C r}_{2} O_{3} \to + 6 C r O_{4}^{2 -} . . . (i)$
Eq wt of $N a_{2} C r O_{4} = \frac{162}{3} g / e q$

$Reduction half : + 5 N O_{3}^{-} \to + 3 N O_{2}^{-} . . . (i i)$
Eq wt of $K N O_{2} = \frac{85}{2} g / e q$

Sum of eq wt: $54 + 42.5 = 96.5 g / e q$

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