See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

2 2 4 4KI HgCl K HgI 2KCl    Remained KI = 5 – 4 = 1 mol 1 mol K2HgI4 and 2 mol KCl is formed. 2 2 4 4 (K HgI 2K HgI )     f o f 1 1 2 2 3 3 ( T ) K (i m i m i m )     o f 0 (T )