Discussion Forum

How long ( approximate ) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66g of diborane?  (  Atomic weight of B = 10.8μ )

Answer:

Explanation:

Given that , i= 100 amp. also 27.66 g of diborane ( B₂H₆)

 Molecular mass of B₂H₆ = $10.8\times 2+6 =27.6$

  Number of moles of B₂H₆  in 27.66 g=  $\frac{Given mass}{Molecular mass}=\frac{27.66}{27.6}=1$

Now consider the equation

  $B_{2}H_{6}+3O_{2}\rightarrow B_{2}O_{3}+3H_{2}O$

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

    $H_{2}O\rightleftharpoons 2H^{+}+O$

From the above  equation it can be easily interpreted that in electrolysis of water for the production of 1 mole of oxygen from 1 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 12 × 96500

   We know Q= it

So,       12 × 96500 = 100 × t in seconds

or           $\frac{12\times 96500}{100\times 3600}=t$ in hours = 3.2 hours