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)

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

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

catya that's not even a b or c