Answer:
percentage yield = 67%
Explanation:
Given data:
Mass of Cu(NO₃)₂ = 15.25 g
Mass of NaOH = 12.75 g
Percentage yield = ?
Solution:
Cu(NO₃)₂ + 2NaOH → Cu(OH)₂ + 2NaNO₃
Moles of Cu(NO₃)₂:
Number of moles = mass/ molar mass
Number of moles = 15.25 g /187.56 g/mol
Number of moles = 0.08 mol
Moles of NaOH :
Number of moles = mass/ molar mass
Number of moles = 12.75 g / 40 g/mol
Number of moles = 0.32 mol
Now we will compare the moles of Cu(OH)₂ with NaOH and Cu(NO₃)₂.
NaOH : Cu(OH)₂
2 : 1
0.32 : 1/2×0.32 = 0.16 mol
Cu(NO₃)₂ : Cu(OH)₂
1 : 1
0.08 : 0.08
The number of moles produced by Cu(NO₃)₂ are less so it will limiting reactant.
Mass of Cu(OH)₂:
Mass = number of moles × molar mass
Mass = 0.08 mol × 97.6 g/mol
Mass = 7.808 g
Theoretical yield = 7.808 g
Percent yield:
percentage yield = Actual yield/ theoretical yield × 100
percentage yield = 5.23 g/ 7.808 g × 100
percentage yield = 0.67 × 100
percentage yield = 67%
