Question

For decolorization of $1$ mole of $KMn{O}_4$ the moles of $H_2{O}_2$ required is:

• A. $1/2$
• B. $3/2$
• C. $5/2$
• D. $7/2$

Answer 1

Answer: (B)

$3/2$

Solution:

$3H_2{O}_2+2K{M}_n{O}_4⟶3O_2+M_n{O}_2+2KOH+2H_{2}O$

Above reaction shows decolotrization $K{M}_n{O}_4$

$2$ moles of $K{M}_n{O}_4$ require $3$ mole of $H_2{O}_2$ to reduce

$\therefore 1$ mole of $K{M}_n{O}_4$ will required $3/2$ moles of $H_2{O}_2$ to get reguced into $M_n{O}_2$

$\therefore$ option $B$ is the correct answer.