Question

Excess of ozone when passed through 1 litre of 1M H2O2 solution the volume of oxygen liberated at STP will be.... a) 11.2 litre b) 22.4 lit c) 5.6 lit d) 44.8 lit

Answer 1

Explanation:

5.6litres

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Answer 2

Answer:

The correct answer is option D

Explanation:

$\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{O}_{3} \rightarrow \mathrm{H}_{2} \mathrm{O}+2 \mathrm{O}_{2}$

$\mathrm{M}$ mole

$M=\frac{\eta}{\text { Vlit }}$

$\eta=M \times V( lit )$

$\eta=1 \times 1$

$\eta=1$ mole

1 mole

$\therefore volume of \mathrm{O}_{2}$ liberated at STP when an excess of $\mathrm{O}_{3}$ is passed though 1 lit of $1 \mathrm{M} \mathrm{H}_{2} \mathrm{O}_{2}$ the solution is $44.8$ lit.