When ferrous oxalte is titrated against $K_{2} C r_{2} O_{7}$ m eq. of $F e^{2 +}, C_{2} O_{4}^{2 -}$ and ${C r}_{2} O_{7}^{2 -}$ in the redox reaction are $X, Y$ and $Z$ respectively, then:-

X = Y

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X + Y = Z

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X + 2 Y = Z

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2 X + 6 Y + 6 Z

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Solution

The correct option is A $X = Y$
The formula of ferous oxalate is $F e C_{2} O_{4}$
And the equation is-
$F e C_{2} O_{4} + K_{2} C r_{2} O_{7} \to F e^{+ 2} + C_{2} O_{4}^{- 2} + C r_{2} O_{7}^{2 -}$
If ferrous oxalte is titrated against $K_{2} {C r}_{2} O_{7}$ m eq. of ${F e}^{2 +}$ , $C_{2} O_{4}^{2 -}$ and $C_{2} O_{7}^{2 -}$ in the redox reaction are X,Y and Z respectively, then X=Y
Hence the correct option is A

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