Nassib S. answered • 09/29/15

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Chemistry Tutor!

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!