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Step-by-Step Solution
Step 1: Write the Combustion Reaction of Diborane
Diborane (B2H6) combusts according to the reaction:
$$\text{B}_2\text{H}_6 + 3\,\text{O}_2 \rightarrow \text{B}_2\text{O}_3 + 3\,\text{H}_2\text{O}$$
From this balanced equation, 1 mole of B2H6 requires 3 moles of O2 for complete combustion.
Step 2: Calculate the Number of Moles of Diborane
The molar mass of diborane (B2H6) is computed as:
$$
2 \times 10.8 + 6 = 21.6 + 6 = 27.6 \text{ g/mol}
$$
Given mass of B2H6 = 27.66 g, the number of moles is approximately:
$$
\frac{27.66}{27.6} \approx 1 \text{ mole}
$$
Step 3: Determine the Required Oxygen Moles
For every 1 mole of B2H6, we need 3 moles of O2:
$$
1 \text{ mole of B}_2\text{H}_6 \;\rightarrow\; 3 \text{ moles of O}_2
$$
Hence, we need 3 moles of O2 in total.
Step 4: Relate Oxygen Production to Electrolysis of Water
The reaction for the electrolysis of water (producing O2) is:
$$
2\,\text{H}_2\text{O} \;(l)\; \rightarrow \;\text{O}_2 \;(g)\; + 4\,\text{H}^+ \;(aq)\; + 4\,e^-
$$
This tells us that producing 1 mole of O2 involves 4 moles of electrons (4 Faradays of charge).
Step 5: Use Faraday’s Laws to Relate Charge and Moles of O₂
• 3 moles of O2 require:
$$
3 \times 4 = 12 \text{ Faradays of charge}
$$
• One Faraday corresponds to 96500 coulombs of charge.
• Therefore, the total charge $Q$ needed to generate 3 moles of O2 is:
$$
Q = 12 \times 96500 \text{ C}
$$
Step 6: Calculate the Required Time
If the current $I = 100$ amperes flows, we can use:
$$
Q = I \times t
$$
Hence,
$$
12 \times 96500 = 100 \times t
$$
$$
t = \frac{12 \times 96500}{100} = 11580 \text{ s}
$$
Step 7: Convert the Time into Hours
$$
t \text{ (in hours)} = \frac{11580}{3600} \approx 3.22 \text{ hours}
$$
This is approximately 3.2 hours.
Answer: 3.2 hours
