H2O2+2KI40% yield⟶I2+2KOHH2O2+2KMnO4+3H2SO450% yield⟶K2SO4+2MnSO4+3O2+4H2O150 mL of a H2O2 sample was divided into two parts. The first part was treated with KI and the formed KOH required 200mL of M2H2SO4 for complete neutralisation. The other part was treated with KMnO4 yielding 6.72 L of O2 at 1 atm and 273 K. Using the % yields indicated, find the volume strength of the H2O2 sample used.

H2O2+2KI40% yield⟶I2+2KOHH2O2+2KMnO4+3H2SO450% yield⟶K2SO4+2MnSO4+

1.	H2O2+2KI40% yield⟶I2+2KOHH2O2+2KMnO4+3H2SO450% yield⟶K2SO4+2MnSO4+3O2+4H2O150 mL of a H2O2 sample was divided into two parts. The first part was treated with KI and the formed KOH required 200mL of M2H2SO4 for complete neutralisation. The other part was treated with KMnO4 yielding 6.72 L of O2 at 1 atm and 273 K. Using the % yields indicated, find the volume strength of the H2O2 sample used.
Answer» $H_{2} O_{2} + 2 K I 40 % y i e l d ⟶ I_{2} + 2 K O H H_{2} O_{2} + 2 K M n O_{4} + 3 H_{2} S O_{4} 50 % y i e l d ⟶ K_{2} S O_{4} + 2 M n S O_{4} + 3 O_{2} + 4 H_{2} O$ $150 mL$ of a $H_{2} O_{2}$ sample was divided into two parts. The first part was treated with $K I$ and the formed $K O H$ required $200 mL$ of $\frac{M}{2} H_{2} S O_{4}$ for complete neutralisation. The other part was treated with $K M n O_{4}$ yielding $6.72 L$ of $O_{2}$ at $1 atm$ and $273 K$ . Using the $%$ yields indicated, find the volume strength of the $H_{2} O_{2}$ sample used.