$N a_{2} C_{2} O_{4}$ and $K H C_{2} O_{4} . H_{2} C_{2} O_{4}$ are mixed in such a proportion by weight that the normality of the resulting solution as a reducing agent is $3.62$ times the normality as an acid. What is the proportion?
[Multiply your answer by 100].

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Solution

Let the proportion by weight of $N a_{2} C_{2} O_{4}$ and $K H C_{2} O_{4} . H_{2} C_{2} O_{4}$ be $x : y$ .
Let $M$ be the molar mass of the mixture.
The equivalent mass as an acid is $\frac{M}{3 y}$ .
The equivalent mass as an reducing agent is $\frac{M}{2 (x + 2 y)}$ .
Hence, $3.62 \times \frac{M}{3 y} = \frac{M}{2 (x + y)}$ .
$7.2 x + 7.2 y = 3 y$
$7.2 x = 4.2 y$
Hence, $\frac{x}{y} = \frac{1.7}{1} = 1.7 \times 100 = 170$

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Na2C2O4 and KHC2O4.H2C2O4 are mixed in such a proportion by weight that the normality of the resulting solution as a reducing agent is 3.62 times the normality as an acid. What is the proportion?[Multiply your answer by 100].

Let the proportion by weight of Na2C2O4 and KHC2O4.H2C2O4 be x:y.Let M be the molar mass of the mixture.The equivalent mass as an acid is M3y.The equivalent mass as an reducing agent is M2(x+2y).Hence, 3.62×M3y=M2(x+y).7.2x+7.2y=3y7.2x=4.2yHence, xy=1.71=1.7×100=170

$N a_{2} C_{2} O_{4}$ and $K H C_{2} O_{4} . H_{2} C_{2} O_{4}$ are mixed in such a proportion by weight that the normality of the resulting solution as a reducing agent is $3.62$ times the normality as an acid. What is the proportion?
[Multiply your answer by 100].