A solution containing 0.714 g of Sn^2+ (Sn 119) as SnCl2 is treated with excess HgCl2 to give Hg2Cl2 (SnCl2 + 2HgCl2 -> SnCl4 + Hg2Cl2). All the Hg2Cl2 is then reduced by further SnCl2 (SnCl2 + Hg2Cl2 -> SnCl4 + 2Hg). Hg = 200. The mass of grey mercury finally formed and the total moles of SnCl2 used are

JEE Advanced levelSingle correct optionDifficulty 3 of 3qc053

A solution containing $0.714 g$ of $Sn X^{2 +}$ (Sn 119) as $SnCl X_{2}$ is treated with excess $HgCl X_{2}$ to give $Hg X_{2} Cl X_{2}$ ( $SnCl X_{2} + 2 HgCl X_{2} SnCl X_{4} + Hg X_{2} Cl X_{2}$ ). All the $Hg X_{2} Cl X_{2}$ is then reduced by further $SnCl X_{2}$ ( $SnCl X_{2} + Hg X_{2} Cl X_{2} SnCl X_{4} + 2 Hg$ ). Hg = 200. The mass of grey mercury finally formed and the total moles of $SnCl X_{2}$ used are

n (Sn X^{2 +}) = 0.714/119 = 0.006 mol

gives

n (Hg X_{2} Cl X_{2}) = 0.006 mol

in the first step. The second step needs another 0.006 mol of

SnCl X_{2}

and yields

2 \times 0.006 = 0.012 mol

Hg

, i.e.

0.012 \times 200 = 2.40 g

. Total

SnCl X_{2} = 2 \times 0.006 = 0.012 mol