Question

Reaction 1.

a) For each mole of Cu dissolved in the nitric acid solution (HNO3), how many moles of [Cu(H2O)6]2+ are formed?

b) Given your data above, how many moles of [Cu(H2O)6]2+ were formed in your reaction?

Reaction I:

Cu (s) + 4 H3O+(aq) + 2 NO3-(aq) --> [Cu(H2O)6]2+(aq) + 2 NO2(g)

The first reaction in the series is an oxidation – reduction reaction where copper metal“dissolves” in nitric acid (HNO3). Anoxidation – reduction reactionoccurs when one atomgains an electron while the other atom loses an electron. The nitrate ion oxidizes the coppermetal to a copper (II) ion, Cu2+, while at the same producing brown NO2(toxic!) gas. The copper(II) ion is soluble in water, so as the reaction proceeds, the metal ribbon disappears. The copper(II) ions in solution bind to six water molecules. The formation of [Cu(H2O)6]2+ion causes the solution to turn blue (Giunta). After all the metallic copper has been oxidized, water is added to dilute the reaction mixture and stop the reaction.

Reaction 2.

a) For each mole of Cu(H2O)62+ that go into Reaction II, how many moles of Cu(OH)2 are formed?

b) In your reaction, how many moles of Cu(OH)2 were formed?

Reaction II:

[Cu(H2O)6]2+(aq) + 2 OH---> Cu(OH)2(s) + 6 H2O (l)

Sodium hydroxide is added to the aqueous solution containing [Cu(H2O)6]2+ions to form theinsoluble Cu(OH)2precipitate; the copper(II) ion reacts with sodium hydroxide in a double displacement (metathesis) precipitation reaction. Precipitation reactions involve the formation of a precipitate (solid) from two aqueous compounds. The hydroxide ion binds to the copper(II) ion more efficiently (stronger) than water does. The hydroxide ion displaces the water molecules from the copper(II) ion, yielding copper hydroxide, a light blue precipitate in a clear solution (solution may appear light blue from the precipitate). Sodium hydroxide, a strong base, will neutralize any of the remaining H3O+ ions from the nitric acid; once the reaction mixture becomes basic, the precipitation of Cu(OH)2 is complete (Giunta).

Reaction 3.

a) In the experiment, for each mole of Cu(OH)2, how many moles of CuO are formed?

b) In your results for the reactions, how many moles of CuO were formed?

Reaction III:

Cu(OH)2(s) --> CuO (s) + H2O (l)

Once the precipitation of Cu(OH)2 is complete, the basic mixture containing the precipitate is heated to decompose the Cu(OH)2. During this decomposition reaction, copper hydroxide is broken down into its two compounds: copper(II) oxide and water. The observable physical change is the conversion of copper hydroxide (Cu(OH)2), the light blue precipitate, into a black, crystalline copper(II) oxide (CuO) when it is heated (Giunta). There are a few decomposition complications involving heat. One of them is that metallic hydroxides, when heated, decompose into a metallic oxide and H2O (Askew). The heat cannot fully decompose the copper hydroxide into two elemental parts, and therefore, the result is two compounds, one of which is copper(II) oxide and the other is water.

Note: I don't know if this is helpful but when I did the experiment the mass of Cu started as 0.192 g and went to 0.511 with 266.1% recovered (it got messed up.)

Answer 1

Reaction-1

(a):Mole ratio between Cu(s) and Cu(H₂O)₆]²⁺(aq) = 1 / 1

Moles of Cu(H₂O)₆]²⁺(aq) formed = 1 mol Cu(s) * [1 mol Cu(H₂O)₆]²⁺(aq) / 1 mol Cu(s)] = 1 mol Cu(H₂O)₆]²⁺(aq) (answer)

Reaction-2:

(a): Mole ratio between Cu(H₂O)₆]²⁺(aq) and Cu(OH)2(s) = 1 / 1

Moles of Cu(OH)₂(s) formed = 1 mol Cu(H₂O)₆]²⁺(aq) * [1 mol Cu(OH)₂(s) / 1 mol Cu(H₂O)₆]²⁺(aq)] = 1 mol Cu(OH)₂(s) (answer)

Reaction-3:

(a): Mole ratio between Cu(OH)2(s) and CuO = 1 / 1

Moles of CuO formed = 1 mol Cu(OH)₂(s) * [1 mol CuO / 1 mol Cu(OH)₂(s)] = 1 mol CuO (answer)