Question

Given the following chemical reaction, what mass of copper (II) hydroxide would be formed from the reaction of 25.5 mL of 1.25M copper (II) sulfate with excess sodium hydroxide?

CuSO₄​(aq)+2NaOH(aq)→Cu(OH)₂​(s)+Na₂​SO₄​(aq)

Answer 1

To find the mass of copper (II) hydroxide formed from 25.5 mL of 1.25M copper (II) sulfate, calculate the moles of copper (II) sulfate, which is 0.031875 moles. Using the 1:1 molar ratio from the balanced equation, you get the same moles of copper (II) hydroxide, which is 3.11 g when multiplied by its molar mass.

Step-by-step explanation:

To calculate the mass of copper (II) hydroxide, Cu(OH)2, from the reaction of 25.5 mL of 1.25M copper (II) sulfate with an excess of sodium hydroxide (NaOH), we first need to determine the number of moles of copper (II) sulfate reacting.

We use the molarity equation:

• Moles of CuSO4 = volume (L) × molarity (M)

Therefore:

• Moles of CuSO4 = 0.0255 L × 1.25 M = 0.031875 moles

From the balanced chemical equation:

CuSO4(aq) + 2NaOH(aq) → Cu(OH)2(s) + Na2SO4(aq)

1 mole of CuSO4 produces 1 mole of Cu(OH)2, thus, we have 0.031875 moles of Cu(OH)2 produced.

The molar mass of Cu(OH)2 is approximately 97.56 g/mol. Therefore:

Mass of Cu(OH)2 = moles × molar mass

Mass of Cu(OH)2 = 0.031875 moles × 97.56 g/mol = 3.11 g (to 3 significant figures)