Question

Given:

$2MnO_4^-(aq.) + 5H_2O_2 + 6H^+ \longrightarrow 2Mn^{2+} + 5O_2 + 8H_2O$

100 ml 0.4M $MnO_4^-$ reacts with 50 ml $H_2O_2$. The volume strength of $H_2O_2$ is.

• A. 11.35 V
• B. 22.7 V
• C. 34.5 V
• D. 44.8 V

Answer 1

Answer: (D)

44.8 V

Answer 2

Solution:

1. Calculate moles of $MnO_4^-$:
   Volume = 100 mL = 0.1 L, Concentration = 0.4 M
   Moles of $MnO_4^-$ = 0.1 × 0.4 = 0.04 mol

2. Determine moles of $H_2O_2$ required:
   From the balanced equation:
   $2 \text{ } MnO_4^- : 5 \text{ } H_2O_2$
   Thus, moles $H_2O_2$ = (5/2) × 0.04 = 0.1 mol

3. Find molarity of the $H_2O_2$ solution:
   Given that 0.1 mol $H_2O_2$ are present in 50 mL = 0.05 L
   Molarity = 0.1 / 0.05 = 2.0 M

4. Relate $H_2O_2$ to $O_2$ production:
   From the equation, 1 mol $H_2O_2$ produces 1 mol $O_2$.
   Thus, 1 L of 2.0 M $H_2O_2$ generates 2.0 mol $O_2$.

5. Calculate volume strength (at NTP):
   At NTP, 1 mol $O_2$ = 22.4 L
   Volume of $O_2$ from 1 L $H_2O_2$ = 2.0 × 22.4 = 44.8 L