Question

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)

Answer 1

2.624 g

Step-by-step 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