Question

The equivalent weight of ${HNO}_3$ in the following reaction is :

$3Cu+8HN{O}_3⟶3Cu{\left({NO}_3\right)}_2+2NO+4H_{2}O$

• A. 63
• B. 189
• C. 126
• D. 84

Answer 1

Answer: (D)

84

The given reaction is,

$3Cu+8HN{O}_3\to 3Cu(N{O}_3{)}_2+2NO+4H_{2}O$

8 moles of $HN{O}_3$ gives- 6moles of $N{O}_3^{-}$ and 2 moles of $NO$

So, $HN{O}_3\to NO$ the oxidation number of $N$ changes from +5 to +2.

$N{O}_3^{-}+3e^{-}\to NO(+5$ to $+2)$

For 2 moles of $NO$. 6 elctrons are there.

But moles of $HN{O}_3$ =8

So equivalent weight$=\frac{MolarMass}{No.of{e}^{-}}\times 8$

$\therefore$ Equivalent weight$=\frac{63\times 8}{6}=84$