H2O2+2KI40% yield−−−−−→I2+2KOHH2O2+2KMnO4+3H2SO450% yield−−−−−→K2SO4+2MnSO4+3O2+4H2O150 mL of H2O2 sample was divided into two parts. First part was treated with KI and formed KOH required 200 mL of M2 H2SO4 for neutralization. Other part was treated with KMnO4 for neutralization yielding 6.74 L of O2 at 1 atm and 273 K. Using % yield indicated, find volume strength of H2O2 sample used:

H2O2+2KI40% yield−−−−−→I2+2KOHH2O2+2KMnO4+3H2SO450% yield�

1.	H2O2+2KI40% yield−−−−−→I2+2KOHH2O2+2KMnO4+3H2SO450% yield−−−−−→K2SO4+2MnSO4+3O2+4H2O150 mL of H2O2 sample was divided into two parts. First part was treated with KI and formed KOH required 200 mL of M2 H2SO4 for neutralization. Other part was treated with KMnO4 for neutralization yielding 6.74 L of O2 at 1 atm and 273 K. Using % yield indicated, find volume strength of H2O2 sample used:
Answer» $H_{2} O_{2} + 2 K I 40 % y i e l d - ---- \to I_{2} + 2 K O H$ $H_{2} O_{2} + 2 K M n O_{4} + 3 H_{2} S O_{4} 50 % y i e l d - ---- \to K_{2} S O_{4} + 2 M n S O_{4} + 3 O_{2} + 4 H_{2} O$ $150 m L$ of $H_{2} O_{2}$ sample was divided into two parts. First part was treated with $K I$ and formed $K O H$ required $200 m L$ of $\frac{M}{2} H_{2} S O_{4}$ for neutralization. Other part was treated with $K M n O_{4}$ for neutralization yielding $6.74 L$ of $O_{2}$ at $1 a t m$ and $273 K$ . Using $%$ yield indicated, find volume strength of $H_{2} O_{2}$ sample used: