Answer:
Percentage yield = 61.7%
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated:
2AgNO₃ + Cu —> Cu(NO₃)₂ + 2Ag
Next, we shall determine the mass of AgNO₃ that reacted and the mass of Ag produced from the balanced equation. This is illustrated below:
Molar mass of AgNO₃ = 108 + 14 + (16×3)
= 108 + 14 + 48
= 170 g/mol
Mass of AgNO₃ from the balanced equation = 2 × 170 = 340 g
Molar mass of Ag = 108 g/mol
Mass of Ag from the balanced equation = 2 × 108 = 216 g
SUMMARY:
From the balanced equation above,
340 g of AgNO₃ reacted to produce 216 g of Ag.
Next, we shall determine the theoretical yield of Ag. This can be obtained as follow:
From the balanced equation above,
340 g of AgNO₃ reacted to produce 216 g of Ag.
Therefore, 51 g of AgNO₃ will react to produce = (51 × 216)/340 = 32.4 g of Ag.
Thus, the theoretical yield of Ag is 32.4 g.
Finally, we shall determine the percentage yield of Ag. This can be obtained as follow:
Actual yield = 20 g
Theoretical yield = 32.4 g
Percentage yield =?
Percentage yield = Actual yield /Theoretical yield × 100
Percentage yield = 20 / 32.4 × 100
Percentage yield = 61.7%
