Answer:
Percent yield = 63.3%
Explanation:
Given data:
Mass of copper = 32.89 g
Mass of oxygen = 32.89 g
Mass of copper(II) oxide formed = 26.20 g
Mass of calcium chloride produced = ?
Solution:
The actual yield of reaction is the amount that you have recovered such as 26.20 g.
Theoretical yield will be calculated in the following way,
Chemical equation:
2Cu + O₂ → 2CuO
Number of moles of Cu:
Number of moles = Mass /molar mass
Number of moles = 32.89 g / 63.55 g/mol
Number of moles = 0.52 mol
Number of moles of O₂:
Number of moles = Mass /molar mass
Number of moles = 32.89 g / 32 g/mol
Number of moles = 1.03 mol
Now we will compare the moles of CuO with Cu and O₂.
O₂ : CuO
1 : 2
1.03 : 2.06
Cu : CuO
2 : 2
0.52 : 0.52
The number of moles of CuO produced by Cu are less so it will limiting reactant.
Mass of CuO: (THEORETICAL YIELD)
Mass = number of moles × molar mass
Mass = 0.52 mol × 79.55 g/mol
Mass = 41.4 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 26.20 g/ 41.4 g × 100
Percent yield = 63.3%
