How many grams of copper (II) hydroxide can be prepared from 2.4 grams of copper (II) nitrate (Cu(NO3)2 ) and excess sodium hydroxide?

percentage yield = 67%

Explanation:

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%

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