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Question

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?
(Given: Atomic weight of B=10.8 u)

A
6.4 hours
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B
3.2 hours
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C
0.8 hours
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D
1.6 hours
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Solution

The correct option is B 3.2 hours
Given that, i=100 amperes also, 27.66 g of diborane (B2H6 )
Molecular mass of B2H6=10.8×2+6=27.6
Number of moles of B2H6 in 27.66 g = Given massMolar mass=27.6627.6=1(approx)
Now consider the equation:
B2H6 + 3O2 B2O3 + 3H2O
From the equation we can interpret that 3 moles of oxygen is required to burn 1 mole (i.e. 27.6 g) B2H6 completely.
Also consider the electrolysis reaction of water i.e.
H2O 2H+ + O2
2H+2HH2
O22O+2eO2

From the above equation it can be easily interpreted that in electrolysis of water for the production of 1 mole of oxygen from 2 mole of H2O at anode 4 moles electrons are required.
Likewise for the production of 3 moles of O2, 12 (3×4) moles of electrons will be needed.
So, the total amount of charge required to produce 3 moles of oxygen will be 12 F or 2×96500
We know; Q=i×t
Given: (i=100 Amperes)
So, = 12×96500=100×t (in seconds)
= 12×96500100×3600=t (in hours)=3.2 hours




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