Question

opper(II) nitrate reacts with sodium hydroxide to produce a precipitate of light blue copper(II) hydroxide. (a) Write the net ionic equation for the reaction. (b) Calculate the maximum mass of copper(II) hydroxide that can be formed when 2.00 g of sodium hydroxide is added to 80.0 mL of 0.500 m Cu(NO3)2(aq)

Answer 1

The answer to your question is:

a) Cu⁺² + 2NO₃⁻¹ + Na⁺¹ + 2OH⁻¹ ⇒ Cu⁺² + 2OH⁻¹ + Na⁺¹ + 2NO₃⁻¹

b) 2.45 g of Cu(OH)₂

Step-by-step explanation:

Data

Copper (II) nitrate + Sodium hydroxide produces copper (II) hydroxide

(a) Write the net ionic equation for the reaction.

Cu(NO₃)₂ + 2NaOH ⇒ Cu(OH)₂ + 2NaNO₃

Cu⁺² + 2NO₃⁻¹ + Na⁺¹ + 2OH⁻¹ ⇒ Cu⁺² + 2OH⁻¹ + Na⁺¹ + 2NO₃⁻¹

(b) Calculate the maximum mass of copper(II) hydroxide that can be formed when 2.00 g of sodium hydroxide is added to 80.0 mL of 0.500 m Cu(NO3)2(aq).

Cu(NO₃)₂ + 2NaOH ⇒ Cu(OH)₂ + 2NaNO₃

MW NaOH = 23 + 16 + 1

= 40 g

40g of NaOH --------------- 1 mol

2 g ------------- x

x = (2 x 1) / 40

x = 0.05 mol

Molarity = moles / volume

moles = Molarity x volume

moles = 0.5 x 0.08

= 0.04 mol

The proportion is 1:2, 1 mol of Cu(NO₃)₂ to 2 mol of NaOH

Then the limiting reactant is NaOH because the proportion is 0.04 to 0.05 is not 1:2.

2 moles of NaOH ----------- 1 mol Cu(OH)₂

0.05 mol -------------------- x

x = (0.05 x 1) / 2

x = 0.025 mol of Cu(OH)₂

MW of Cu(OH)₂ = 64 + 32 + 2 = 98 g

98 g of Cu(OH)₂ --------------- 1 mol

x --------------- 0.025 mol

x = (0.025 x 98) / 1

x = 2.45 g