Question 4
(a) Starting with calcium chloride, describe briefly how a solid sample of calcium
trioxocarbonate (IV) can be prepared in the laboratory.
[5 marks]
(b) With relevant equations outline the procedure for the purification of impure copper
[7 marks]
(c) Copper reacts with concentrated trioxonitrate (V) acid:
(i) write a balanced chemical equation for the reaction;
(ii) state what would be observed in the reaction;
(iii) state why the copper is oxidized
(iv) an excess of copper is added to 25.0cm3 of 16.0 mol dm-3 HNO3. Calculate the
volumes of the gas formed at s.t.p.
[H= 1.0, N = 14.0, O = 16.0, Cu = 63.0; Molar volume of gas at s.t.p. = 22.4 dm3]
[9 marks]
(d) (i) pure HNO3 is a colourless liquid but when exposed to air, it turns yellowish-
brown in colour. Explain briefly this observation.
(ii) Write a balanced equation for the laboratory preparation of hydrogen trioxonitrate (V) acid.
[4 marks]
Observation
This question was not popular among the candidates as majority of them avoided it. Few candidates that responded to it performed below average.
In part (a), majority of the candidates could state how a solid sample of calcium trioxocarbonate (IV) can be prepared in the laboratory by starting with calcium chloride.
In part (b), majority of the candidates could not outline the procedure for the purification of impure copper using relevant equations.
In part (c), majority of the candidates could not write a balanced chemical equation for the reaction of copper with concentrated trioxonitrate (V) acid. In addition to this, they could not state what would be observed in the reaction.
In part (d), few candidates stated the brown colour of NO2, but failed to say it dissolved in the acid to cause the colouration.
The expected answers include:
- (a) Water is added to the CaCl2 to form a solution. Na2CO3 is then added to the
Solution to precipitate CaCO3 which is filtered, washed and dried.
(b) Electricity is passed through a solution of CuSO4 using an impure copper as the
Anode and pure copper as the cathode.
During electrolysis, the anode loses mass as copper dissolves and the cathode gains mass as
copper is deposited.
Anode: Cu(s) → Cu2+(aq) + 2e-
Cathode: Cu2+(aq) + 2e- → Cu(s)
(c) (i) Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
(ii) - blue solution
- (reddish) brown gas
- gas bubbles / effervescence
- metal dissolves / metal disappears
(iii) Because copper loses electrons / oxidation number increases
(iv) Moles of acid = 0.025 x 16
= 0.4 mole
Moles of NO2 = moles of acid
2
= 0.2 mole
Volume of gas (NO2) at s.t.p. = 0.2 x 22.4 dm3 = 4.48 dm3
Alternative
C1V1 = C2V2
16.0 x 25 = C2 x 1000
C2 = 16.0 * 251000
= 0.40 mol
4 mol HNO3 = 63g Cu
0.4 mol = 6.3g Cu
63g Cu = 2 x 22.4 dm3 NO2
6.3 g Cu = 44.863 x 6.3
= 4.48 dm3
(d) (i) Pure HNO3 undergoes decomposition to NO2 (brown gas) (1). This gas dissolves (1) in the rest
of the acid leaving it with that colour
(ii) - 2 KNO3 + H2SO4 → K2SO4 +2 HNO3
- NaNO3 + H2SO4 → NaHSO4 + HNO3
- 2NaNO3 + H2SO4 → Na2SO4 + 2HNO3
- KNO3 + H2SO4 → KHSO4 + HNO3