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22 December, 05:57

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?

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  1. 22 December, 06:32
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    percentage yield = 67%

    Explanation:

    Given dа ta:

    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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