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
Answer : The correct option is, (B) Salicylic acid
Solution :
First we have to calculate the moles of salicylic acid and acetic anhydride.
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 moles of salicylic acid react with 1 mole of acetic anhydride
So, 0.507 moles of salicylic acid react with 
mole of acetic anhydride
The excess of acetic anhydride = 0.783 - 0.2535 = 0.5295 moles
That means the in the given balanced reaction, salicylic acid is a limiting reagent because it limits the formation of products and acetic anhydride is an excess reagent.
Hence, the limiting reagent is, salicylic acid.
There are two iron atoms and three oxygen atoms in each molecule of Fe2O3. So that's five atoms per molecule.
One mole of any molecular substance contains <span>6.02×1023</span> molecules. Since there are two iron atoms in each of the molecules we're considering, there will be <span><span>(6.02×1023)</span>⋅2=1.204×1024</span> iron atoms in a whole mole of them.
But we're considering 0.550 mol, so multiply by 0.550:
<span>n=(0.550</span> mol<span>)⋅<span>(3.01×1024</span></span> atoms/mol<span>)=6.62×1023</span> iron atoms.
