27 grams of Silver was reacted with excess sulfur, according to the following equation: 2Ag+S ---> Ag2S
25 grams of silver sulfide was collected, what is the the theoretical yield, actual yield, and percent yield? Please explain or show your work, it'll really help.

percent yield = 80.7%

Theoretical yield of silver sulfide = 30.975 g

Actual yield  = 25 g

Explanation:

Given data:

Mass of silver = 27 g

Mass of silver sulfide formed = 25 g

Theoretical yield of silver sulfide = ?

Actual yield = ?

Percent yield = ?

Solution:

Chemical equation:

2Ag + S   →     Ag₂S

Number of moles of Ag:

Number of moles = mass/molar mass

Molar mass of silver = 107.8 g/mol

Number of moles = 27 g/ 107.8 g/mol

Number of moles = 0.25 mol

Now we will compare the moles of silver with silver sulfide.

               Ag         :       Ag₂S

               2            :         1

              0.25        :      1/2×0.25 = 0.125 mol

Theoretical yield of silver sulfide:

Mass = number of moles × molar mass

Mass = 0.125 mol ×  247.8 g/mol

Mass = 30.975 g

Actual yield is given in question = 25 g

Percent yield:

Percent yield = (actual yield / theoretical yield) × 100

Percent yield = (25 g/ 30.975 g ) × 100

percent yield = 80.7%

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