What volume of O2 measured at standard conditions will be formed by the action of 100 mL of 0.5NKMnO4 on hydrogen peroxide in an acidic solution?
A
0.12 L
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.28 L
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.56 L
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
1.12 L
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B0.56 L The reaction is as follows: 2KMnO4+5H2O2+3H2SO4→2MnSO4+5O2+K2SO4+8H2O So, number of moles of KMnO4=0.1 (given) So, 100×0.11000=0.01 mol of KMnO4 will produce (52)×0.01=0.025 mol of O2=0.025×22.4=0.56 L.