1. Write the combustion reaction of diborane

Diborane burns in oxygen as:

$\mathrm{B_2H_6 + 3O_2 \rightarrow B_2O_3 + 3H_2O}$

So, 1 mole of diborane requires 3 moles of oxygen.

2. Calculate moles of diborane

Molar mass of diborane:

$M(\mathrm{B_2H_6}) = 2(10.8) + 6(1) = 21.6 + 6 = 27.6\,\text{g mol}^{-1}$

Given mass = $27.66\,\text{g}$

$n(\mathrm{B_2H_6}) = \frac{27.66}{27.6} \approx 1.002\,\text{mol} \approx 1\,\text{mol}$

Thus oxygen needed:

$n(O_2) = 3 \times 1.002 \approx 3.006\,\text{mol} \approx 3\,\text{mol}$

3. Relate oxygen evolved in electrolysis of water to charge passed

At anode:

$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$

So, 1 mole of $O_2$ requires 4 moles of electrons.

Therefore for $3.006$ mol $O_2$:

$n(e^-) = 4 \times 3.006 = 12.024\,\text{mol e}^-$

Charge required:

$Q = n(e^-)F = 12.024 \times 96500 \approx 1.160 \times 10^6\,\text{C}$

4. Calculate time using current

Given current:

$I = 100\,\text{A} = 100\,\text{C s}^{-1}$

Using

$Q = It \Rightarrow t = \frac{Q}{I}$

$t = \frac{1.160 \times 10^6}{100} = 1.160 \times 10^4\,\text{s}$

Convert to hours:

$t = \frac{11600}{3600} \approx 3.22\,\text{h}$

So the approximate time is:

$\boxed{3.2\,\text{hours}}$

5. Option check

• A: $1.6$ h ❌
• B: $6.4$ h ❌
• C: $0.8$ h ❌
• D: $3.2$ h ✅

Hence, the correct answer is D.