Respuesta :
Answer:
Percent yield = 22.8 %
Explanation:
Step 1: Data given
Numbers of moles copper = 0.0970 moles
Mass of copper(I) sulfide = 1.76 grams
Step 2: The balanced equation
2Cu + S ⇒ Cu2S
Step 3: Calculate moles of Cu2S
For 2 moles Cu we need 1 mol S to produce 1 mol Cu2S
For 0.0970 moles Cu we'll hace 0.0970 / 2 = 0.0485 moles
Step 4: Calculate mass of Cu2S
Mass Cu2s = moles Cu2S * molar mass Cu2S
Mass Cu2S = 0.0485 moles * 159.16 g/mol
Mass Cu2S = 7.72 grams
Step 5: Calculate percent yield
Percent yield = (actual yield/ theoretical mass) * 100%
Percent yield = (1.76 grams / 7.72 grams)*100%
Percent yield = 22.8 %
The percentage yield of the experiment obtained by the reaction of 0.0970 mole of copper with excess sulfurs is 22.8%
We'll begin by calculating the number of mole of Cu₂S produced from the reaction. This can be obtained as follow:
2Cu + S —> Cu₂S
From the balanced equation above,
2 moles of Cu reacted to produce 1 mole of Cu₂S.
Therefore,
0.0970 mole of Cu will react to produce = [tex]\frac{0.0970}{2}[/tex] = 0.0485 mole of Cu₂S.
Next, we shall determine the theoretical yield by calculating the mass of 0.0485 mole of Cu₂S.
Molar mass of Cu₂S = (63.5×2) + 32 = 159 g/mol
Mole of Cu₂S = 0.0485 mole
Mass of Cu₂S =?
Mass = mole × molar mass
Mass of Cu₂S = 0.0485 × 159
Mass of Cu₂S = 7.7115 g
Thus, the theoretical yield of Cu₂S is 7.7115 g
Finally, we shall determine the percentage yield of Cu₂S.
Actual yield = 1.76 g
Theoretical yield = 7.7115 g
Percentage yield =?
[tex]Percentage yield = \frac{Actual}{Theoretical} * 100\\\\= \frac{1.76}{7.7115} * 100\\\\[/tex]
Percentage yield = 22.8%
Therefore, the percentage yield of the experiment is 22.8%
Learn more: https://brainly.com/question/10950696