Benzene can be nitrated with a mixture of nitric and sulfuric acids. How do we do that?
1 answer:
Answer:
We can do the nitration of benzene by treating the benzene with a mixture of nitric acid and sulphuric acid by not extending the temperature of 50°C
Explanation:
Nitration of benzene takes place by treating the benzene with a mixture of nitric acid and sulphuric acid at low temperatures such as the temperatures below 50°C
The nitration of benzene takes place through electrophilic substitution reaction
In this reaction the electrophile is nitronium ion (NO2+) which performs an electrophilic substitution reaction on the benzene ring and during the reaction an intermediate will also be formed in which there will be positive charge distributed in the benzene
These electrophile is generated when nitric acid is treated with sulphuric acid
As nitric acid is a strong oxidising agent, here in this case the oxidation state of nitrogen will change from +5 to +3
The reactions regarding the nitration of benzene is present in the file attached
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-20.16 KJ of heat are released by the reaction of 25.0 g of Na2O2.
Explanation:
Given:
mass of Na2O2 = 25 grams
atomic mass of Na2O2 = 78 gram/mole
number of mole =
=
=0. 32 moles
The balanced equation for the reaction:
2 Na2O2(s) + 2 H2O(l) → 4 NaOH(aq) + O2(g) ∆Hο = −126 kJ
It can be seen that 126 KJ of energy is released when 2 moles of Na2O2 undergoes reaction.
similarly 0.3 moles of Na2O2 on reaction would give:
=
x =
= -20.16 KJ
Thus, - 20.16 KJ of energy will be released.
Answer : The concentration of is,
Explanation :
When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is and
is excess reagent.
First we have to calculate the moles of KSCN.
Moles of KSCN = Moles of = Moles of
=
Now we have to calculate the concentration of
Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L
Thus, the concentration of is,