JEE PYQ: P Block Elements Question 29
Question 29 - 2020 (05 Sep Shift 2)
Reaction of ammonia with excess $\text{Cl}_2$ gives:
(1) $\text{NH}_4\text{Cl}$ and $\text{N}_2$
(2) $\text{NH}_4\text{Cl}$ and HCl
(3) $\text{NCl}_3$ and $\text{NH}_4\text{Cl}$
(4) $\text{NCl}_3$ and HCl
Show Answer
Answer: (4)
Solution
$\text{NH}_3 + 3\text{Cl}_2 (\text{excess}) \to \text{NCl}_3 + 3\text{HCl}$. If $\text{NH}_3$ is used in excess then $\text{NH}_4\text{Cl}$ is formed instead of $\text{NCl}_3$.
BETA
