How long (in hours) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.6 g of diborane? (Atomic wt. of B = 10.8 u)

$B_2{H}_6+3O_2\to B_2{O}_3+3H_{2}O$ 
molar mass of  $B_2{H}_6=(2\times 10.8)+(1\times 6)=27.6$
$\therefore$  Mole of  $B_2{H}_6$  in  $27.66g=\frac{27.66}{27.6}\simeq 1$
Thus 1 mole  $B_2{H}_6$  requires 3 mole  $O_2$.
In electrolysis of  $H_{2}O$
Anode :  $2H_{2}O\to 4H^{+}+O_2+4e$
$\therefore n-\text{factor\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}}{O}_2=4$
    Cathode :  $2H_{2}O+2e\to 2O{H}^{-}+H_2]\times 2$   .
Redox          $6H_{2}O\to 4H^{+}+4O{H}^{-}+2H_2+O_2$
or                 $2H_{2}O\to 2H_2+O_2$
Now equivalent of $O_2=n\times$ mole of  $O_2$
                                    $=4\times 3=12$
Also                  $\frac{w}{E}=\frac{i\cdot t}{96500}$
$\therefore t=\frac{w}{E}\times \frac{96500}{i}=\frac{12\times 96500}{100}\sec$
 $t=\frac{12\times 96500}{100\times 3600}hr=3.216hr$