Question

What is the equivalent weight of $N{H}_3$ in the given reaction
$3CuO+2N{H}_3\to 3Cu+N_2+3H_{2}O$

• A. $17$
• B. $\frac{17}{4}$
• C. $\frac{17}{2}$
• D. $\frac{17}{3}$

Answer 1

Answer: (D)

$2N^{3-}\to (N_2{)}^{\circ }+6e^{-}$
2 mole of $N{H}_3$ = 1 mole $N_2$
Thus equivalents
$2\times n=1\times 6$
$n=6/2=3$
$\therefore$ Eq. wt. $=\frac{M}{3}=\frac{17}{3}$