Question

4. (a) Starting with calcium chloride, describe briefly how a solid sample of calcium trioxocarbonate (IV) can be prepared in the laboratory.
(b) With relevant equations, outline the procedure for the purification of impure copper.
(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 25.0 cm3 of 16.0 mol dm⁻³ HNO3. Calculate the volume 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]

Accepted Answer

(a) Water is added to CaCl2 to form a solution. Na2CO3 solution is then added to precipitate CaCO3, which is filtered, washed and dried.

(b) Electricity is passed through a solution of CuSO4, using impure copper as the anode and pure copper as the cathode. During electrolysis, the anode loses mass as copper dissolves into solution, and the cathode gains mass as copper is deposited on it.
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) Observations: a blue solution forms; a reddish-brown gas is evolved; gas bubbles/effervescence occurs; the metal dissolves/disappears.
(iii) The copper is oxidized because it loses electrons — its oxidation number increases (from 0 to +2).
(iv) Moles of acid = 0.025 dm3 × 16 mol dm⁻³ = 0.4 mol. Moles of NO2 = (moles of acid)/2 = 0.2 mol (from the equation, 4 mol HNO3 produces 2 mol NO2). Volume of NO2 at s.t.p. = 0.2 mol × 22.4 dm3/mol = 0.48 dm3.
(Cross-check via the copper side: 4 mol HNO3 reacts with 1 mol Cu = 63 g; 0.4 mol HNO3 reacts with 6.3 g Cu, consistent with excess copper being present; the gas volume computed above, 0.48 dm3, is the answer.)

WAEC Chemistry 2023 Theory — Question 5

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