Question

The following conversion can be obtained by using

$\underset{\text{ (excess) }}{{\mathrm{NH}}_3}\stackrel{?}{\to }{\mathrm{N}}_2{\mathrm{H}}_4$

• A. $OC{l}^{-}$
• B. $HS{O}_3^{-}$
• C. $HC{O}_3^{-}$
• D. $P{O}_4^{-3}$

Answer 1

Answer: (A)

$2\underset{\text{excess}}{{\text{NH}}_{\text{3}}}\text{+}\text{NaOCl}\to {\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{+}\text{NaCl}\text{+}{\text{H}}_{\text{2}}\text{O}$