Question

In a reaction excess of H2O2 is added to 0.1 mole of acidified KMnO4 solution.Then the stp Volume of o2 liberated is:​

Answer 1

Answer:

Explanation:

No of gram equivalent = no. of moles (n) × nf = normality × volume (In litres)

Kmn04 is an oxiding agent and will oxide O- of H2O2 to O2 (oxidation state=0). Hence nf of H202 is 2.

==> 2 × n(H2O2) = 0.4 × 0.5

==> n = 0.1

Hence 0.1 mole of H2O2 is required.

Answer 2

Answer:

The correct answer is $5.6$ liter.

Explanation:

A volume exists simply described as the quantity of area populated by any three-dimensional solid. These solids can be a cube, a cuboid, a cone, a cylinder, or a sphere.

Step 1

In a reaction excess of $\mathrm{H}_{2} \mathrm{O}_{2}$ is added to $0.1$ mole of acidified KMnO4 solution.

Equivalents of $\mathrm{H}_{2} \mathrm{O}_{2}$ needed to react completely with $0.1$ mole

$\mathrm{kMnO}_{4}=0.1 \times \mathrm{nf}\left(\mathrm{n}_{\mathrm{f}}\right. of \left.\mathrm{kMnO}=5\right)$

$=0.5$

Step 2

Now, the chemical reaction be

$\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}+1 / 2 \mathrm{O}_{2}$

$1$ equivalent $\mathrm{H}_{2} \mathrm{O}_{2}$ produce

$\frac{22-4}{2}=11 \cdot 2 \mathrm{~L} \mathrm{O}_{2}$

$0.5$ equivalent $\mathrm{H}_{2} \mathrm{O}_{2}$ will produce $0.5 \times 11 \cdot 2=5.6 \mathrm{~L} \mathrm{O}_{2}$

Therefore, the correct answer is $5.6$ liter.

#SPJ2