How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane? (Atomic weight of B = 10.8 u)

First we find how many moles of diborane, $$\mathrm{B_2H_6}$$, are to be burnt. Its molar mass is obtained by adding the atomic masses of all the atoms present:

$$M_{\mathrm{B_2H_6}} \;=\;2\times M_{\mathrm B}+6\times M_{\mathrm H} \;=\;2\times10.8\;\text{u}+6\times1\;\text{u} \;=\;21.6\;\text{u}+6\;\text{u} \;=\;27.6\;\text{g mol}^{-1}.$$

The given mass of diborane is $$27.66\;\text{g}$$, so its amount in moles is

$$n_{\mathrm{B_2H_6}} =\frac{27.66\;\text{g}}{27.6\;\text{g mol}^{-1}} \approx1.002\;\text{mol}\; \approx1\;\text{mol}.$$

Now we write the complete combustion reaction of diborane with oxygen:

$$\mathrm{B_2H_6+3\,O_2\;\longrightarrow\;B_2O_3+3\,H_2O.}$$

The balanced equation shows that one mole of $$\mathrm{B_2H_6}$$ requires three moles of $$\mathrm{O_2}$$. Therefore the moles of oxygen needed are

$$n_{\mathrm{O_2}}=3\times n_{\mathrm{B_2H_6}} =3\times1\;\text{mol}=3\;\text{mol}.$$

Next we relate the electricity passed in electrolysis of water to the amount of oxygen liberated. In the electrolysis of water the cathodic and anodic processes together can be summarized as

$$2\,\mathrm{H_2O}\;\longrightarrow\;\mathrm{O_2}+4\,\mathrm{H^+}+4\,e^-.$$

This half-reaction shows that four moles of electrons (4 F) produce one mole of $$\mathrm{O_2}$$. Faraday’s law of electrolysis states

$$Q = n_e\,F,$$

where $$Q$$ is the total charge passed, $$n_e$$ is the total moles of electrons involved, and $$F=96500\;\text{C mol}^{-1}$$ is the Faraday constant.

For each mole of $$\mathrm{O_2}$$ the charge required is

$$Q_{1\;\mathrm{mol\;O_2}} = 4F = 4 \times 96500\;\text{C}=386000\;\text{C}.$$

Therefore, to obtain $$3$$ moles of $$\mathrm{O_2}$$ we must supply

$$Q = 3\times386000\;\text{C}=1\,158\,000\;\text{C}.$$

The current given is $$I = 100\;\text{A} = 100\;\text{C s}^{-1}$$. Using the relation

$$t=\frac{Q}{I},$$

the time required becomes

$$t=\frac{1\,158\,000\;\text{C}}{100\;\text{C s}^{-1}} =11\,580\;\text{s}.$$

To convert seconds into hours we divide by the number of seconds in one hour, $$3600\;\text{s}$$:

$$t=\frac{11\,580}{3600}\;\text{h}\approx3.2167\;\text{h}\approx3.2\;\text{h}.$$

Hence, the correct answer is Option D.