Fundamental M9 Clarification

[nicole_ershaghi 3F]

Can anyone explain how to answer the following problem:

Copper(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).

[chloe lee]

Hi,

For part a, use the given facts to write the net ionic equation. Copper(II) nitrate reacts with sodium hydroxide to produce a precipitate of light blue copper(II) hydroxide, so the two reactants are Cu(NO3)2 and NaOH.
you could write out all of the aqueous components in the solution:
Cu2+ (aq) + NO3- (aq) + Na+ (aq) + OH- (aq) --> Cu(OH)2 (s) + NO3- (aq) + OH- (aq)
after cancelling out spectator ions you get:
Cu2+ (aq) + OH- (aq) --> Cu(OH)2 (s)
balanced:
Cu2+ (aq) + 2OH- (aq) --> Cu(OH)2 (s)

For part b, we would use the given 2.00 g NaOH and 80.0 mL of 0.500 M Cu(NO3)2 to find the limiting reactant and then see what amount of Cu(OH)2 could be formed.
(2.00 g NaOH) (1 mol NaOH / 40.00 g NaOH) (1 mol OH- / 1 mol NaOH) (1 mol Cu(OH)2 / 2 mol OH-) = 0.025 mol Cu(OH)2
(80.0 mL) (1 L/10^3 mL) (0.500 mol Cu(NO3)2 / 1L) (1 mol Cu2+ / 1 mol Cu(NO3)2) (1 mol Cu(OH)2 / 1 mol Cu2+) = 0.04 mol Cu(OH)2
0.025 mol Cu(OH)2 < 0.04 mol Cu(OH)2 so the NaOH is limiting reactant (it produces less product)
Next, we take the 0.025 mol Cu(OH)2 and convert to grams.
(0.025 mol Cu(OH)2) (97.57 g Cu(OH)2 / 1 mol Cu(OH)2) = 2.44 g Cu(OH)2
So, the answer to part b is that the maximum mass that can be formed is 2.44 g Cu(OH)2.