Excess alkaline K2HgI4 reacts with 10.2 mg of NH3 according to 2K2HgI4 + NH3 + 3KOH -> HgO.Hg(NH2)I + 7KI + 2H2O. Atomic masses: Hg 200, I 127, N 14, O 16, H 1. The mass of brown precipitate formed is

JEE Advanced levelSingle correct optionDifficulty 3 of 3qc147

Excess alkaline $K X_{2} HgI X_{4}$ reacts with 10.2 mg of $NH X_{3}$ according to $2 K X_{2} HgI X_{4} + NH X_{3} + 3 KOH HgO \cdot Hg (NH X_{2}) I + 7 KI + 2 H X_{2} O$ . Atomic masses: Hg 200, I 127, N 14, O 16, H 1. The mass of brown precipitate formed is

Moles of

NH X_{3}

= 10.2/17 = 0.60 mmol, and 1 mol gives 1 mol of

HgO \cdot Hg (NH X_{2}) I

. Its molar mass = 2(200)+16+14+2+127 = 559 g/mol. Mass = 0.6 mmol x 559 = 335.4 mg, i.e. about 335 mg.