Calculate the mass of anhydrous oxalic acid, which can be oxidised to CO 2 ( g ) by 100 ml of an MnO − 4 solution, 10 ml of which is capable of oxidising 50 ml of 1 N I − to I 2 .

Question

Calculate the mass of anhydrous oxalic acid, which can be oxidised to CO 2 ​ ( g ) by 100 ml of an MnO − 4 ​ solution, 10 ml of which is capable of oxidising 50 ml of 1 N I − to I 2 ​ .

TestHub · Accepted Answer

Solution:

First, determine the normality of the MnO₄⁻ solution.

For the reaction between MnO₄⁻ and I⁻:

$N_{1} V_{1} (M n O_{4}^{-}) = N_{2} V_{2} (I^{-})$

$N_{1} \times 10 m l = 1 N \times 50 m l$

$N_{1} = 5 N$

Now, for the reaction between MnO₄⁻ and oxalic acid (H₂C₂O₄):

$N_{1} V_{1} (M n O_{4}^{-}) = N_{2} V_{2} (o x a l i c a c i d)$

$N_{2} V_{2} = \frac{W}{E} \times 1000$ (where W is mass, E is equivalent weight)

$5 N \times 100 m l = \frac{W}{E _{o x a l i c a c i d}} \times 1000$

For oxalic acid (H₂C₂O₄), the oxidation state of carbon changes from +3 to +4 (in CO₂). Each carbon loses 1 electron, so for C₂ it loses 2 electrons.

Molecular weight of H₂C₂O₄ = $2 (1) + 2 (12) + 4 (16) = 2 + 24 + 64 = 90 g / m o l$

Equivalent weight $E_{o x a l i c a c i d} = \frac{M o l ec u l a r w e i g h t}{n - f a c t or} = \frac{90}{2} = 45 g / e q$

Substitute E into the equation:

$5 \times 100 = \frac{W}{45} \times 1000$

$500 = \frac{1000 W}{45}$

$W = \frac{500 \times 45}{1000}$

$W = \frac{1 \times 45}{2}$

$W = 22.5 g m$

The final answer is $22.5 g m$ .

Chemistry - Redox Reactions Question with Solution | TestHub

Options:

Doubts & Discussion

JEE Main (2028)

JEE Main + Advanced (1 year 12th class)