Question:
The oxide that gives $\mathrm{H}_{2} \mathrm{O}_{2}$ most readily o treatment with $\mathrm{H}_{2} \mathrm{O}$ is :
Correct Option:
Solution:
1. $\mathrm{PbO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Pb}(\mathrm{OH})_{4}$
2. $\mathrm{Na}_{2} \mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_{2} \mathrm{O}_{2}$
this reaction is possible at room temperature
3. $\mathrm{SnO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Sn}(\mathrm{OH})_{4}$
4. Acidified $\mathrm{BaO}_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}$ gives $\mathrm{H}_{2} \mathrm{O}_{2}$ after evaporation.
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