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?
(Atomic weight of B $=$ 10.8 u)

• A. 1.6 hours
• B. 6.4 hours
• C. 0.8 hours
• D. 3.2 hours

Answer 1

Answer: (D)

$B_2{H}_6+3O_2\to B_2{O}_3+3H_{2}O$
According to balanced equation
27.66 g $B_2{H}_6$ i.e. 1 mole $B_2{H}_6$ requires 3 mole of $O_2$ . Now this oxygen is produced by electrolysis of water as follows
$2H_{2}O\stackrel{4F}{\to }2H_2+O_2$
As 1 mole $O_2$ is produced by 4 F charge
So 3 mole $O_2$ will be produced by 12 F charge
So now on applying
Q $=$ It
$\text{12}\times \text{965}00C=100\times t(s)$
$t=\frac{12\times 96500}{100\times 3600}$ hours
$t=3.2$ hours