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11 February, 07:20

What volume (ml) of 3.0 m naoh is required to react with 0.8024-g copper (ii) nitrate? what mass of copper (ii) hydroxide will form, assuming 100% yield?

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  1. 11 February, 08:46
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    The reaction between NaOH and Cu (NO₃) ₂ is as follows

    2NaOH + Cu (NO₃) ₂ - - - > 2NaNO₃ + Cu (OH) ₂

    Q1)

    stoichiometry of NaOH to Cu (NO₃) ₂ is 2:1

    this means that 2 mol of NaOH reacts with 1 mol of Cu (NO₃) ₂

    the mass of Cu (NO₃) ₂ reacted - 0.8024 g

    molar mass of Cu (NO₃) ₂ is 187.56 g/mol

    therefore the number of Cu (NO₃) ₂ moles that have reacted

    - 0.8024 g / 187.56 g/mol = 0.00427 mol

    according to the stoichiometry, number of NaOH moles - 0.00427 mol x 2

    then number of NaOH moles that have reacted - 0.00855 mol

    In a 3.0 M NaOH solution, 3 moles are in 1000 mL of solution

    Then volume required for 0.00855 mol - 1000 x 0.00855 / 3 = 2.85 mL

    2.85 mL of 3.0 M NaOH is required for this reaction

    Q2)

    Assuming that there's 100 % yield of Cu (OH) ₂, we can directly calculate the mass of Cu (OH) ₂ formed from the number of moles of reactants that were used up.

    Stoichiometry of Cu (NO₃) ₂ to Cu (OH) ₂ is 1:1

    this means that 1 mol of Cu (NO₃) ₂ gives a yield of 1 mol of Cu (OH) ₂

    the number of Cu (NO₃) ₂ moles that reacted - 0.00427 mol

    Therefore an equal amount of moles of Cu (OH) ₂ were formed

    Then amount of Cu (OH) ₂ moles produced - 0.00427 mol

    Mass of Cu (OH) ₂ formed - 0.00427 mol x 97.56 g/mol = 0.42 g

    A mass of 0.42 g of Cu (OH) ₂ was formed in this reaction
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