The oxide that gives hydrogen peroxide $(H_{2} O_{2})$ on the treatment with a dilute acid $(H_{2} {S O}_{4})$ is

{M n O}_{2}

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{P b O}_{2}

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N a_{2} O_{2}

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{T i O}_{2}

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Solution

The correct option is C $N a_{2} O_{2}$
$N a_{2} O_{2}$ upon reacting with dilute sulphuric acid gives hydrogen peroxide and sodium sulphate.
The chemical reaction is as follows :
$N a_{2} O_{2} + H_{2} S O_{4} \to N a_{2} S O_{4} + H_{2} O_{2}$

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Similar questions

The oxide that gives hydrogen peroxide ( $H_{2} O_{2}$ ) on the treatment with a dilute acid ( $H_{2} S O_{4}$ ) is

H_{2} O_{2}

H_{2} O_{2}

K_{2} {C r}_{2} O_{7} + K I + H_{2} {S O}_{4} \to K_{2} {S O}_{4} + {C r}_{2} {({S O}_{4})}_{3} + I_{2} + H_{2} O

K M n O_{4} + {N a}_{2} {S O}_{3} \to M n O_{2} + {N a}_{2} {S O}_{4} + K O H

C u + H {N O}_{3} \to C u {({N O}_{3})}_{2} + {N O}_{2} + H_{2} O

K M {n O}_{4} + H_{2} C_{2} O_{4} + H_{2} {S O}_{4} \to K_{2} {S O}_{4} + {M n S O}_{4} + {C O}_{2} + H_{2} O

Question 7

The oxide that gives $H_{2} O_{2}$ on treatment with dilute $H_{2} S O_{4}$ is

(a) $P b O_{2}$ (b) $B a O_{2} .8 H_{2} O + P_{2}$

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