PLEASE HELP!!!? Percent composition problem?

Question

Silver is reacted with excess mercury (II) chloride, producing a solid silver chloride precipitate. If the actual yield of 25.5 g silver chloride represents a 77 percent yield, how much silver was initially present?

Ag (s) + HgCl2 (aq) ------> AgCl (s) + Hg (l)

I already balanced the equation:

2 Ag + HgCl2 -------> 2 AgCl + Hg

I am just so confused on how to solve this!
What I know for sure is that the percent yield is actual yield/theoretical yield * (100), but I am not sure if that formula even matters in this question. I would love to learn though!
Thank you for your help!

Answer 1

2 Ag + HgCl2 -------> 2 AgCl + Hg
If 25.5 g of AgCl represents 77% yield, then you can calculate what 100% yield would be. That is 25.5/77 = x/100 and
x = 33.1 g
How many moles of AgCl is this?
33.1 g AgCl x 1 mol/143.3 g = 0.231 moles AgCl
From the balanced equation, the mole ratio of Ag : AgCl is 1: 1
Therefore, moles Ag present initially = 0.231 (this could be the answer, if you want it in moles). If you want the mass...
0.231 moles Ag x 107.9 g/mol = 24.9 g