I could really use some help with this chemistry percent yield 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% 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

Hey Angelica,

 

For these types of problems, you're usually asked to calculate the percent yield. Since you're given the actual yield (25.5g) and the percent yield (77%), you're going to work backwards and calculate the theoretical yield and then use that number to determine the amount of silver (Ag) initially used. 

 

Since % yield= (actual/theoretical), then 0.77=(25.5g / theoretical) --> ~33.1g

 

This is the amount of silver chloride (AgCl) which means we can now use this value and work backwards to find the amount of silver (Ag) via stoichiometry. 

 

Using the molar mass and mole to mole ratios of AgCl and Ag and the balanced equation...

 

33.1g AgCl * (1 mol AgCl / 143.32g ) * (2mol Ag / 2mol AgCl) * (107.87g / 1mol Ag) = ~24.9 grams of Ag

 

Hope this helps!