# 200 ml solution of H_2O_2 completely oxidises 125g  Fe^+. Find out the molarity and volume strength of H_2O_2?

May 28, 2017

$\left[{H}_{2} {O}_{2}\right] = 5.6 M$ and Vol Stgth (g/L) = 190 g/L

#### Explanation:

In an acid solution
${H}_{2} {O}_{2} + 2 F {e}^{2 +} + 2 {H}^{+}$ => $2 F {e}^{3 +} + 2 {H}_{2} O$

From the reaction ratios between ${H}_{2} {O}_{2}$ and $F {e}^{2 +}$

2 moles ${H}_{2} {O}_{2}$ = 1 mole $F {e}^{2 +}$ = 1 mole $F {e}^{3 +}$

2((0.20L)[${H}_{2} {O}_{2}$] = ((125g(Fe^(3+)))/((56"g/(mol))Fe^(3+)))) = $2.23 m o l F {e}^{3 +}$

$\left[{H}_{2} {O}_{2}\right] = \left(\frac{2.23}{\left(2\right) \left(0.20\right)}\right) M o l a r {H}_{2} {O}_{2} = 5.60 M \text{in} {H}_{2} {O}_{2}$

$V o l . \text{Strength} \left(\frac{g}{L}\right)$ = $\left(\frac{\left(5.6 \text{mol} {H}_{2} {O}_{2}\right)}{L}\right) \left(0.200 L\right) = 1.12 m o l {H}_{2} {O}_{2}$ = $38 g {H}_{2} {O}_{2} \text{in} 200 m l$ => $0.190 g \frac{{H}_{2} {O}_{2}}{m l}$=> ((190gH_2O_2))/L)