Answer:
percent yield = 80.7%
Theoretical yield of silver sulfide = 30.975 g
Actual yield = 25 g
Step-by-step 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%