Question

The volume of 6.8v of H2O2 required to liberate 224 ml of O2 at STP IS

Answer 1

concept :- there are lot of methods to show strength of hydrogen peroxide. one of most famous method is $\textbf{as number volume}$ : the most common practice of expressing the concentration of hydrogen peroxide solution is in term of volume of oxygen which a solution of hydrogen peroxide gives on decomposition by heat. for example, 10 volume hydrogen peroxide refers to a solution of hydrogen peroxide whose, 1 litre will give 10 litre of oxygen at STP.

a/c to question, 6.8 volume hydrogen peroxide. means, 1 litre of hydrogen peroxide oxides gives 6.8 litres of oxygen at STP.

reaction is $2H_2O_2\rightarrow 2H_2O+O_2$

here it is clear that, 22.4L of oxygen is obtained from $H_2O_2$ = 2 × 34 = 68g

6.8L of oxygen is obtained from $H_2O_2$ = 68/22.4 × 6.8 = 20.64g
hence, strength of hydrogen peroxide is 20.64g/L

now, volume of hydrogen peroxide required to liberate 224ml.
mole of oxygen = 224/22400 = 1/100
so, mole of hydrogen peroxide = 2/100 = 0.02
now, weight of hydrogen peroxide = 0.02 × 68 = 1.36g

so, volume of hydrogen peroxide = 1.36/20.64
= 0.0658914729 L
= 658ml

Answer 2

Answer:

Refer to the given picture

Explanation: