Question

$KMn{O}_4$ reacts with oxalic acid according to the equation:
$2Mn{O}_4{}^{-}+5C_2{O}_4^{2-}+16H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$

20 mL of 0.1 M $KMn{O}_4$ will react with?

• A. 120 mL of 0.25 M oxalic acid
• B. 150 mL of 0.1 M oxalic acid
• C. 50 mL of 0.1 M oxalic acid
• D. 150 mL of 0.2 M oxalic acid

Answer 1

The correct option is C 50 mL of 0.1 M oxalic acid

$2Mn{O}_4{}^{-}+5C_2{O}_4^{2-}+16H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$

$2\times 22.4LatNTP$ $5\times 22.4LatNTP$

$\because 2\times 22.4L$ reacts with $5\times 22.4L$ of oxalic acid.

$\therefore 20ml$ of $0.1M$ $K{M}_n{O}_4$ reacts with oxalic acid, $=\frac{5\times 22.4\times 20}{2\times 22.4}ml=50ml$

So, The crrect option is $50ml$ of $0.1M$ oxalic acid.