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)

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