Stannous chloride reduces mercuric chloride first to $Hg X_{2} Cl X_{2}$ and then (with excess) to metallic mercury. How many moles of $SnCl X_{2}$ are needed to convert 0.20 mol of $HgCl X_{2}$ completely to metallic Hg?

Answer: B

Step 1:

2 HgCl X_{2} + SnCl X_{2} Hg X_{2} Cl X_{2} + SnCl X_{4}

uses 0.10 mol

SnCl X_{2}

and forms 0.10 mol

Hg X_{2} Cl X_{2}

. Step 2:

Hg X_{2} Cl X_{2} + SnCl X_{2} 2 Hg + SnCl X_{4}

uses another 0.10 mol. Total = 0.20 mol; overall each Hg(II) takes 2 electrons from one Sn(II).