Question

Copper(II) nitrate reacts with sodium hydroxide to produce a precipitate of light blue copper(II) hydroxide. a) Write a net ionic equation for this reaction. b) Identify the spectator ions or ions. c) Calculate the maximum mass of copper(II) hydroxide that can be formed when 2g of sodium hydroxide is added to 80 mLof 0.500 M Cu(NO3)2 (aq).

Answer 1

a) Cu(NO₃)₂ + 2NaOH Cu(OH)₂ + 2NaNO₃

Cu²⁺ + 2 NO₃^- + 2Na⁺ + 2OH^- Cu²⁺ + 2OH^- + 2Na⁺ + 2 NO₃^-

b)Here Na+ and NO3^- are spectator ions as they do not participate in the reaction.

c) We can wee from the equation for 1 mole of Cu(NO3)2, 2 moles of NaOH is required

0.5 M Cu(NO3)2, means 0.5 moles in 1L of solution,

therefore in 80 mL , i.e. 0.080L moles of Cu(NO3)2 = 0.08 X 0.5 = 0.04 moles

so NaOH required for complete consumption of Cu(NO3)2 = 0.04 X 2 = 0.08 moles

Given, 2 g of NaOH, therefore moles of NaOH = mass/molar mass = 2/40 = 0.05 moles

so NaOH is limiting reagent and Cu(NO3)2 is in excess

So amount of Cu(OH)2 formed will depend on amount of NaOH.

as we can see from the equation 2 moles of NaOH gives 1 mol of Cu(OH)2

so, 0.05 moles of NaOH will give 0.025 moles of Cu(OH)2

so maximum mass of Cu(OH)2 can be formed = no. of moles X molar mass = 0.025 X 97.5 = 2.4375g