J.R. S. answered • 09/28/15

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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