Given: $H_{2} O_{2} + {KMnO}_{4} + H_{2} {SO}_{4} \to {MnSO}_{4} + H_{2} O + O_{2} + K_{2} {SO}_{4}$ .Which of the following correct statement(s)?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Number of gm equivalent of $H_{2} O_{2}$ and ${KMnO}_{4}$ reacts in the ratio of 5: 2

Number of moles of ${MnSO}_{4}$ and $O_{2}$ produced in the ratio of 2: 5

Equivalent mass of $H_{2} {SO}_{4}$ is 29.4

$100 ml 2.27 {V H}_{2} O_{2}$ reacts completely with $16 ml 0.5 {M KMnO}_{4}$ in given reaction

answer is B, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$5 \times (H_{2} O_{2} \to 2 H^{'} + O_{2} + 2 e) 2 \times ({MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} + {Mn}^{2 +} + 4 H_{2} O) 5 H_{2} O_{2} + 2 {KMnO}_{4} + 3 H_{2} {SO}_{4} \to 5 O_{2} + 2 {MnSO}_{4} + 8 H_{2} O + K_{2} {SO}_{4}$
A) No. of gm equivalent of $H_{2} O_{2} =$ no. of gm equivalent of ${KMnO}_{4}$
B) $\frac{{Moles of MnSO}_{4}}{{moles of O}_{2}} = \frac{2}{5}$
C) $E = \frac{98}{10} \times 3 = 29.4$
D) $x_{H_{2} O_{2}} = x_{{KMnO}_{4}} \frac{2 \times 100 \times 0.2}{1000} = \frac{5 \times 0.5 V}{1000} \Rightarrow V = 16 ml$