Question

Given,\[34.1g\] of \[Pb{O_4}\] is dissolved in \[500mL\] of \[4M\] \[HN{O_3}\], then (atomic weight of \[Pb = 206amu)\] -
A.There is no residue
B.The weight of residue is \[11.2g\]
C.\[300mL\]of \[6M\] $NaOH$ is required to neutralize excess of \[HN{O_3}\]
D.Total \[2.2\] moles of $NaOH$ are used after reaction of \[P{b_3}{O_4}\] with \[HN{O_3}\]

Answer 1

Hint: We need to know that the lead tetroxide is a chemical compound having the formula, \[P{b_3}{O_4}\] which is also known as minimum or red lead. It is a bright orange colour compound and it is used for the production of rust proof primer paints and batteries. The nitric acid is a strong acid having the chemical compound, \[HN{O_3}\] and it is a highly corrosive mineral acid. By the decomposition of nitric acid, there is a formation of water and oxides of nitrogen. The nitric acid is used for the preparation of ammonium nitrate for fertilizers.

Complete answer:
The residue is produced in this reaction. Hence, option (A) is incorrect.
The residue present in the mixture is equal to lead nitrate and lead nitrate.
Number of moles of water formed during the reaction \[ = 2 \times 0.0497 = 0.0994\]moles
Hence, mass of water formed \[ = 18 \times 0.094 = 1.79g\]
The molar mass of nitric acid \[ = 63g/mol\]
Then mass of nitric acid consumed is equal to \[12.53g\] and mass of residue is equal to \[44.8g\]. Hence, option (B) is incorrect.
When one mole of lead tetroxide is reacted with four moles of nitric acid, there is a formation of one mole of lead oxide with two moles of lead nitrate and two moles of water. Let’s see the reaction,
\[P{b_3}{O_4} + 4HN{O_3} \to Pb{O_2} + 2Pb{\left( {N{O_3}} \right)_2} + 2{H_2}O\]
The molecular mass of lead tetroxide \[ = 685.6g/mol\]
\[34.1g\] of lead tetroxide corresponds to \[\dfrac{{34.1}}{{685.6}} = 0.0497moles\].
And it will react with \[4 \times 0.0497 = 0.1989molesofHN{O_3}\]
Out of \[4 \times 0.5 = 2\,\] moles of nitric acid, \[2 - 0.1989 = 1.80\] moles of nitric acid will be neutralized with \[1.80\] moles of sodium hydroxide.
Here, \[1.80\] moles of sodium hydroxide corresponds to \[300mL\] of \[6M\] sodium hydroxide. Hence, option (C) is correct.
The number of $NaOH$ is used after the reaction of \[P{b_3}{O_4}\] with \[HN{O_3}\] is not equal to \[2.2\]moles. Hence, option (D) is incorrect.

Hence, option (C) is correct.

Note:
We have to know that when the lead tetroxide is reacted with concentrated nitric acid, there is a formation of lead oxide and lead nitrate with water molecules. And this reaction may be obtained by the evaporation of metallic lead with nitric acid. When the lead is added to the dilute nitric acid, there occurs a redox reaction. If the concentrated nitric acid reacts with lead, there is a formation of lead oxide.