# Question #0a72f

Feb 7, 2017

By definition y vol ${H}_{2} {O}_{2}$ means that sample solution of ${H}_{2} {O}_{2}$ from each ml of which yml ${O}_{2}$ at STP is produced on thermal decomposition.

So y ml ${H}_{2} {O}_{2}$ solution will produce ${y}^{2}$ ml ${O}_{2}$ at STP.

Now the balanced equation of thermal decomposition reaction of ${H}_{2} {O}_{2}$ is as follows

$2 {H}_{2} {O}_{2} \text{ "" "->" "" } {O}_{2} + 2 {H}_{2} O$

$= 2 \text{ "mol" "" "=22400mL" ""at STP}$

This equation reveals that 22400mL ${O}_{2}$ at STP is produced from 2mol ${H}_{2} {O}_{2}$

So ${y}^{2}$ mL ${O}_{2}$ will be produced from $\frac{2 {y}^{2}}{22400}$ mole ${H}_{2} {O}_{2}$ and this amount of ${H}_{2} {O}_{2}$ is present in $y m L$ solution.

Again x mL y(M) $K M n {O}_{4}$ solution will contain $\left(x y \times {10}^{-} 3\right)$ mole $K M n {O}_{4}$

Now the balanced redox reaction is

$5 {H}_{2} {O}_{2} + 2 K M n {O}_{4} + 3 {H}_{2} S {O}_{4} \to {K}_{2} S {O}_{4} + 8 {H}_{2} O + 4 {O}_{2}$ comparing the mole ratio of reactants calculated in the reacting solutions with that of balanced redox reaction we can write

$\frac{x y \times {10}^{-} 3}{\frac{2 {y}^{2}}{22400}} = \frac{2}{5}$

$\implies \frac{x}{y} = \frac{0.4}{11.2} \approx 0.036$