Required reaction :

B₂H₆ + 3O₂ $$ \to $$ B₂ O₃ + 3 H₂ O

Here molar mass of B₂H₆ =10.8 $$ \times $$ 2 + 6 = 27.6 gm

Given weight of B₂H₆ = 27.66 g

$$\therefore\,\,\,\,$$No of moles of B₂H₆ = $${{27.6} \over {27.66}} \simeq 1$$ mole.

For combustion of 1 mole B₂H₆ 3 moles O₂ required.

This 3 mole of O₂ is obtained by electrolysis of H₂O.

2H₂O($$l$$) $$ \to $$ O₂ (g) + 4 H⁺ (aq) + 4 e^$$-$$

From Faradays law of electrolysis,

moles $$ \times $$ n_f = $${{It} \over {96500}}$$

Here moles of O₂ = 3.

N_f of O₂ = 4 (in H₂ change of O = $$-$$2

and in O₂ change of 0 = O.

So change in charge = 2 .

for two atoms of O₂ change in charge = 2 $$ \times $$ 2 = 4)

$$\therefore\,\,\,\,$$ 3 $$ \times $$ 4 = $${{100 \times t} \over {96500}}$$

$$ \Rightarrow \,\,\,\,$$ t = 12 $$ \times $$ 965 sec.

$$ \Rightarrow \,\,\,\,$$ t = $${{12 \times 965} \over {60 \times 60}}$$ hr

= 3.2 hr