To solve this problem, we need to calculate the amount of oxygen produced through the electrolysis of water, and then determine if it's sufficient to completely burn 27.66 g of diborane (B₂H₆). Here's the step-by-step solution:
Determine the stoichiometry of the reaction between diborane and oxygen:
2 B₂H₆ + 6 O₂ → B₂O₃ + 6 H₂O
From the balanced equation, we can see that 2 moles of diborane react with 6 moles of oxygen to produce 2 moles of boron oxide (B₂O₃).
Calculate the molar mass of diborane (B₂H₆):
Boron (B) atomic mass = 10.8 u
Hydrogen (H) atomic mass = 1.01 u
Molar mass of B₂H₆ = 2 * (10.8 u) + 6 * (1.01 u) = 21.62 u
Calculate the number of moles of diborane (B₂H₆) corresponding to 27.66 g:
Number of moles = Mass / Molar mass
Number of moles = 27.66 g / 21.62 g/mol ≈ 1.28 mol
According to the stoichiometry, 2 moles of diborane react with 6 moles of oxygen.
Thus, the number of moles of oxygen needed = 6/2 * number of moles of diborane = 3 * 1.28 mol = 3.84 mol
Now, let's calculate the amount of oxygen produced by the electrolysis of water:
The Faraday's constant (F) is 96,485 C/mol, which is the charge required to deposit one mole of any substance during electrolysis.
Given that the charge passed is 100 A (Amperes) for a certain time (t), the total charge passed (Q) can be calculated using the formula:
Q = I * t
Where:
Q = Charge passed in Coulombs (C)
I = Current in Amperes (A)
t = Time in seconds (s)
Given that 1 Faraday (F) = 96,485 C, and we need 3.84 moles of oxygen, we can find the total charge required:
Q = 3.84 moles * 96,485 C/mol
Now, we can find the time (t) required for this charge to pass with a current of 100 A:
t = Q / I
Plugging in the values:
t = (3.84 * 96,485 C) / 100 A ≈ 3696 seconds
Converting seconds to hours:
t ≈ 3696 s / 3600 s/h ≈ 1.027 hours
So, the closest answer from the given options is (B) 1.6 hours.
