When copper (ii) nitrate is mixed with sodium hydroxide, a blue precipitate, (copper (ii) hydroxide) forms. the other product is sodium nitrate. when 15.25 grams of copper (ii) nitrate is reacted with 12.75 grams of sodium hydroxide, what is the percent yield of the precipitate if 5.23 grams are recovered?

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%

i think slash and