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Question: What mass of \( \;HN{O_3} \) ​ is required to make \( 1{{ }}litre \) of \( 2N \) solution to be used...

What mass of   HNO3\;HN{O_3} ​ is required to make 1litre1{{ }}litre of 2N2N solution to be used as an oxidizing agent in the reaction?
3Cu+8HNO33Cu(NO3)2+2NO+4H2O3Cu + 8HN{O_3}\, \to 3Cu{(N{O_3})_2}\, + 2NO + 4{H_2}O
(A) 63g63g
(B) 21g21g
(C) 42g42g
(D) 84g84g

Explanation

Solution

The chemical name of   HNO3\;HN{O_3} is Nitric acid. It is a strong mineral acid that is also known as spirit of niter and aqua forts. It is highly corrosive and toxic in nature. If it comes in contact with skin, it can cause severe burns on it.

Complete Step-by-step Solution:
The reaction of Copper with nitric acid is an example of redox reaction. A redox reaction is given as a chemical reaction which is governed via a transfer of electrons between two atoms or molecules participating in the reaction. This transfer of electrons is observed by calculating the changes that occur in the oxidation states of the reacting species.
In Redox reactions both oxidation and reduction take place in the same reaction. In this type of chemical reaction, one species is oxidised and the other is reduced.
The molecule in which the addition of electrons takes place gets reduced, and the other from which the removal of electrons takes place gets oxidized.
Now, the given reaction is;
3Cu+8HNO33Cu(NO3)2+2NO+4H2O3Cu + 8HN{O_3}\, \to 3Cu{(N{O_3})_2}\, + 2NO + 4{H_2}O
+5+5+2+ 5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 5\,\,\,\,\,\,\,\, + 2
In this reaction, the oxidation state of nitrogen changes from +5+ 5 to +2+ 2 , and the oxidation state of copper changes from 00 to +2+ 2 . Hence the species, whose oxidation state decreases is said to be reduced, and whose oxidation state increase is said to be oxidised.
Now, we will calculate the mass of   HNO3\;HN{O_3} ​ that is required to make 1 litre1{\text{ }}litre of 2N2N solution. The change in oxidation of NN is 33 . Hence the equivalent weight of   HNO3\;HN{O_3} will be
Equivalentweight=633=21gEquivalent{{ }}weight{{ = }}\dfrac{{63}}{3} = 21g
Hence, for making 1N1N solution we need 21g21g\, of   HNO3\;HN{O_3}
And for making 2N2N we will need 2×21=42g2 \times 21 = 42g of   HNO3\;HN{O_3} .
Therefore, option (C) is correct.

Note:
Oxidation numbers represent the potential charge present on an atom in its ionic form. If the atom's oxidation is decreased in a reaction, then it is said to get reduced. If it increases, then the atom is said to get oxidized