How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66g of diborane?

B2H6 + 3O2 → B2O3 + 3H2O
Moles of O2 required = 3 × moles of B2H6
Molecular weight of B2H6 = 2(10.8) + 6(1) = 27.6 g/mol
Moles of B2H6 = 27.66 g / 27.6 g/mol = 1 mol
Moles of O2 required = 3 × 1 mol = 3 mol
Oxygen is produced by the electrolysis of water:
2H2O → 2H2 + O2
From the stoichiometry, 1 mole of O2 is produced from 4 moles of electrons.
Thus, 3 moles of O2 require 12 moles of electrons.
Charge = moles of electrons × Faraday's constant = 12 mol × 96500 C/mol = 1158000 C
Current = 100 A
Time (t) = Charge / Current = 1158000 C / 100 A = 11580 s
Time in hours = 11580 s / 3600 s/hr ≈ 3.2 hours
So, the correct option is 3.2 hours