Question

The number of coulombs required to reduce 12.3 g of nitrobenzene to aniline is

Answer 1

Answer:

Explanation:n this reaction , Reaction for reduction

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C₆H₅NO₂ +6H⁺ +6e⁻ -----→ C₆H₅NH₂ +2H₂O

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Given in the question :-⇒

Molar mass of nitrobenzene =123

Here 12.3 nitrobenezene Mole= 6×12.3 / 123

= 0.6 

Charge = 0.6 × 96500 

=57900

As 50% efficiency 

So, 2 × 57900

=115800 C .

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Answer 2

The number of coulombs required to reduce 12.3 g of nitrobenzene to aniline is 57900.

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$    $\text{Number of moles of nitrobenzene}=\frac{12.3g}{123g/mol}=0.1mol$  

The reduction of nitrobenzene is represented by:

$C_6H_5NO_2+6H^++6e^\rightarrow C_6H_5NH_2+2H_2O$

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ number of particles.

We know that:

Charge on 1 electron = $1.6\times 10^{-19}C$

Charge on 1 mole of electrons = $1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C$

Charge on 6 mole of electrons = $6\times 96500C=579000C$

1 mole of nitro benzene is reduced by =  579000 C

Thus 0.1 moles of nitro benzene is reduced by =$\frac{ 579000}{1}\times 0.1=57900C$

Hence number of coulombs required to reduce 12.3 g of nitrobenzene to aniline is 57900 C

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