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Question: The equivalent weight of \(HN{O_3}\) (molecular weight = \(63\)) in the following reaction is: \(3...

The equivalent weight of HNO3HN{O_3} (molecular weight = 6363) in the following reaction is:
3Cu+8HNO3→3Cu(NO3)2+2NO+4H2O3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O
A. 4×633\dfrac{4 \times 63}{3}
B. 635\dfrac{{63}}{5}
C. 633\dfrac{{63}}{3}
D. 638\dfrac{{63}}{8}

Explanation

Solution

Equivalent weight which is also known as gram equivalent is the mass of one equivalent that is the mass of a given substance which combines with or displaces a fixed quantity of some other substance. The equivalent weight of an element is the mass which combines with or displaces 1.008 gram of hydrogen or 8.0 grams of oxygen or 35.5 grams of chlorine.

Complete step by step answer: The mathematical formula of equivalent weight is given by:
Equivalent weight = E=MwnE = \dfrac{{{M_w}}}{n}
Where, Mw={M_w} = Molecular weight
n=n = n-factor or valency or the number of electrons transferred
The reaction of copper with nitric acid produces copper nitrate along with nitrous oxide with the release of water molecules. The reaction is as follows:
3Cu+8HNO3→3Cu(NO3)2+2NO+4H2O3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O
In this case, there is a transfer of 2 moles of nitrate ions and these two moles of nitrate ions undergo reduction. The reduction of nitrate ion into nitrous oxide is as follows:
NO3−+3e−→NONO_3^ - + 3{e^ - } \to NO , where n=3n = 3
Overall, there are two nitrate ions and there is a transfer of 6 electrons.
The oxidation state of nitrogen in the NO3−NO_3^ - ion is:
x−6=−1x - 6 = - 1
∴,x=5\therefore ,x = 5
The oxidation state of nitrogen in nitrous oxide is:
x−2=0x - 2 = 0
∴,x=2\therefore ,x = 2
Thus, with the reduction of nitrate ion into nitrous oxide, 3 electrons are required.
Overall, 6 moles of nitrate ions are used up to form 3 moles of copper nitrate
Hence, the equivalent weight of nitric acid is:
E=M+M3=4M3E = M + \dfrac{M}{3} = \dfrac{{4M}}{3}
Thus, E=4×633E = \dfrac{{4 \times 63}}{3} .
Hence, the correct option is A. 4×633\dfrac{{4 \times 63}}{3}.

Note: In a redox equivalent reaction, where the same compound undergoes reduction and oxidation simultaneously, the n-factor plays a crucial role in determining the equivalent weight of the compound undergoing redox reaction