Answer to Question #216979 in Chemistry for Jordy

Question

What mass of copper(II) hydroxide would be produced when a solution containing 6.8g of OH- ions reacts with a solution containing excess Cu2+ ions?

Answer 1

Solution:

copper(II) hydroxide = Cu(OH)₂

Cu²⁺ + 2OH^- → Cu(OH)₂

According to the equation above: n(Cu²⁺) = n(OH^-)/2 = n(Cu(OH)₂)

Calculate the moles of OH^- ions:

(6.8 g OH^-) × (1 mol OH^- / 17 g OH^-) = 0.4 mol OH^-

Thus,

n(Cu(OH)₂) = n(OH^-)/2 = 0.4 mol / 2 = 0.2 mol

Calculate the mass of Cu(OH)₂:

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

Therefore,

(0.2 mol Cu(OH)₂) × (97.56 g Cu(OH)₂ / 1 mol Cu(OH)₂) = 19.512 g Cu(OH)₂ = 19.5 g Cu(OH)₂

Mass of Cu(OH)₂ is 19.5 g.

Answer: 19.5 grams of copper(II) hydroxide (Cu(OH)₂) would be produced.