What is the approximate volume (in mL ) of 10 vol H 2 O 2 solution that will react completely with 1 L of 0.02 M KMnO 4 solution in acidic medium?

Question

What is the approximate volume (in mL ) of 10 vol H 2 O 2 solution that will react completely with 1 L of 0.02 M KMnO 4 solution in acidic medium? (A) 56.05 (B) 113.5 (C) 90.8 (D) 75.75

Tardigrade Admin · Accepted Answer

The balanced equation between the reaction of

K M n O_{4}

and

H_{2} O_{2}

in acidic medium is as follows:

2 K M n O_{4} + 5 H_{2} O_{2} + 3 H_{2} S O_{4} ⟶ 5 O_{2} + 2 M n S O_{4} + K_{2} S O_{4} + 8 H_{2} O

Thus,

2

moles of

K M n O_{4} \equiv 5

moles of

H_{2} O_{2} \equiv 5

moles of

O_{2} (g)

at

STP

.

According to above relation:

34 g

of

H_{2} O_{2}

will give

= 11.2 L

of

O_{2}

at

STP

.

∴ (34 \times 5)

, i.e.

170 g

of

H_{2} O_{2}

will give

= \frac{11.2 \times 170}{34}

= 56 L

∵ 1 L

of

H_{2} O_{2}

give

10 L

of

O_{2}

at

STP

.

∴ 1 m L

of

H_{2} O_{2}

give

10 m L

of

O_{2}

at

STP

.

and

56 m L

of

H_{2} O_{2}

will give

560 m L

of

O_{2}

∴

Volume of

H_{2} O_{2} = 56 m L

Q. What is the approximate volume (in mL ) of 10volH2​O2​ solution that will react completely with 1L of 0.02MKMnO4​ solution in acidic medium?

56.05

113.5

90.8

75.75

Q. What is the approximate volume (in $m L$ ) of $10 v o l H_{2} O_{2}$ solution that will react completely with $1 L$ of $0.02 M K M n O_{4}$ solution in acidic medium?