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)

$B_{2}H_{6}+3O_{2} \rightarrow B_{2}O_{3}+3H_{2}O$
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
$2H_{2}O\overset{4 F}{ \rightarrow }2H_{2}+O_{2}$
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
$\text{12}\times \text{965}00C=100\times t\left(\right.s\left.\right)$
$t=\frac{12 \times 96500}{100 \times 3600}$ hours
$t=3.2$ hours