Question

The volume of octane required to
be used for its combustion by the
oxygen liberated during
electrolysis of NaNO3(aq) by
passing 9.65 ampare current for
1hr. is (consider STP conditions)​

Answer 1

To find:

The volume of octane required to be used for its combustion by the oxygen liberated during

electrolysis of NaNO3(aq) by passing 9.65 ampare current for 1hr.

Calculation:

Number of Faraday's generated

$\rm{ = \dfrac{It}{96500} = \dfrac{9.65 \times (60 \times 60)}{96500} = 0.36F }$

Now , the reaction at Anode:

$\boxed{\sf{2H_{2}O \rightarrow O_{2}+4{H}^{+}+ 4{e}^{-}}}$

So, 4F electricity generates 1 mole of Oxygen.

=> 4F generates 22.4 L of Oxygen.

=> 0.36F generates (22.4/4)(0.36) = 2.016 L of Oxygen.

Now , let's look at the reaction between Octane and Oxygen (Combustion)

$\boxed{\sf{2C_{8}H_{18}+25O_{2}\rightarrow 16CO_{2}+18H_{2}O}}$

So, (25 × 22.4 L) oxygen requires 2 × 22.4 L octane at STP

=> 2.016 L oxygen would require

$= \dfrac{25 \times 22.4}{2 \times 22.4} \times 2.016$

$= \dfrac{25 }{2} \times 2.016$

$= 12.5 \times 2.016$

$= 25.2 \: \rm{litres \: of \: octane \: at \:STP }$

So, final answer is:

$\boxed{ \bf{25.2 \: litres \: of \: octane \: at \:STP }}$

Answer 2

Answer:

The volume of octane required to be used for its combustion by the oxygen liberated during electrolysis of NaNO3(aq) by passing 9.65 ampare current for 1hr. is (consider STP conditions)​ is 25.2 Litres.

Explanation:

• An octane is defined as the measurement of the ability of any fuel that is knocking in when it ignited on a mixture of air in a cylinder of the internal combustion of the engine.
• NaNO3 is known absurdum nitrate. It is an inorganic nitrate of the salt which is related to the sodium.
• It acts as fertilizer.
• It is a solid solution which is seen to be present in a crystalline of the solid nature.

Given that:

An electrolysis of NaNO3 in aqueous which passes through 9.65 ampere current in 1 hour.

To find:

The volume of octane required when using the combustion of the oxygen =?

Formula to be using:

The number of Faraday’s which is generated is:

It/ 96500

=9.65 * (60*60) / 96500

= 9.65* 1200/96500

= 0.36 F

Now, we are going to consider the reaction of the anode:

We get,

2H2o --> O2 +4H- +4e-

Here, 4F is the electricity that is generated by 1 mole of the oxygen.

Oxygen generated is 22.4 L of oxygen

Now, for 0.36 F it gives: (22.4/4) (0.36)

                                     = 2.016 L of oxygen

To find the reaction between octane and oxygen when in combustion

2C8H18 + 25O2 ---> 16CO2 +18H2o

Then, 25* 22.4L of the oxygen will require 2* 22.4 L of the octane at the STP  is 2.016 L of the oxygen

Now we are going to find the value of the STP:

25*22.4/2*22.4 * 2.016

= 25/2 *2.016

= 12.5* 2.016

=25.2 liters of octane is at STP.

#SPJ2