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31 August, 18:26

How many grams of Cu (OH) 2 will precipitate when excess NaOH solution is added to 46.0 mL of 0.584 M CuSO4

solution?

CuSO4 (aq) + 2NaOH (aq) - >Cu (OH) 2 (s) + Na2SO4 (aq)

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  1. 31 August, 18:41
    0
    2.624 g

    Explanation:

    The equation for the reaction is given as;

    CuSO₄ (aq) + 2NaOH (aq) → Cu (OH) ₂ (s) + Na₂SO₄ (aq) Volume of CuSO₄ as 46.0 mL; Molarity of CuSO₄ as 0.584 M

    We are required to calculate the mass of Cu (OH) ₂ precipitated

    We are going to use the following steps; Step 1: Calculate the number of moles of CuSO₄ used

    Molarity = Number of moles : Volume

    To get the number of moles;

    Moles = Molarity * volume

    = 0.584 M * 0.046 L

    = 0.0269 moles

    Step 2: Calculate the number of moles of Cu (OH) ₂ produced From the equation 1 mole of CuSO₄ reacts to give out 1 mole of Cu (OH) ₂ Therefore; Mole ratio of CuSO₄ to Cu (OH) ₂ is 1 : 1.

    Thus, Moles of CuSO₄ = Moles of Cu (OH) ₂

    Hence, moles of Cu (OH) ₂ = 0.0269 moles

    Step 3: Calculate the mass of Cu (OH) ₂

    To get mass we multiply the number of moles with the molar mass.

    Mass = Moles * Molar mass

    Molar mass of Cu (OH) ₂ is 97.561 g/mol

    Therefore;

    Mass of Cu (OH) ₂ = 0.0269 moles * 97.561 g/mol

    = 2.624 g

    Thus, the mass of Cu (OH) ₂ that will precipitate is 2.624 g
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