Question 3 of 7: Pyrometallurgical Processes (20 marks)

Question 3 — Pyrometallurgical Processes (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a)–(e) are conceptual; part (f) is a calculation and follows the Given/Find/steps shape below it.

(a) What is roasting? Roasting is the thermal treatment of a sulphide ore or concentrate in a controlled gas atmosphere, at a temperature high enough to react but below the melting point of the charge, so the solid does not fuse or agglomerate. Its purpose is to change the chemical form of the metal value — typically sulphide to oxide, sulphate or chloride — ahead of smelting, leaching or further pyrometallurgical treatment, and to drive off volatile impurities (As, Sb, Hg) as a gas that can be captured.

(b) Two examples of roasting, with balanced reactions. Oxidising (sulphating-adjacent) roasts of two common sulphides:

Both convert the sulphide to an oxide calcine that can be fed to a reduction (blast or reverberatory) furnace, releasing SO₂ that is normally captured for sulphuric acid manufacture on environmental grounds.

(c) Dead roasting. Dead roasting is carried to completion: enough air and residence time are supplied that essentially all of the sulphide converts to oxide, leaving negligible residual sulphide or sulphate in the calcine. It is used ahead of a purely reductive smelting step (e.g. iron-ore-style reduction to metal, or zinc oxide reduction) where any leftover sulphur would contaminate the metal or the slag. This is the deliberate opposite of a partial "sweetening" roast, which stops short of completion to leave enough sulphide behind to sustain a subsequent matte-smelting step.

(d) Chloridizing roasting. Chloridizing roasting adds a chloridizing agent — commonly NaCl or CaCl₂ — to the charge during roasting so that the metal value is converted to a volatile or water-soluble chloride rather than an oxide. The chloride can then be recovered either by volatilisation (collected from the roaster off-gas) or by aqueous leaching of the calcine, which is useful for metals whose oxides are refractory or difficult to leach directly.

(e) Example of chloridizing roasting. Roasting sphalerite concentrate with salt converts the zinc to a water-soluble chloride while the sulphur reports as sodium sulphate:

The zinc chloride is subsequently recovered by leaching the roasted calcine with water, ahead of electrowinning or precipitation.

(f) ZnS roasted with 25 % excess air — roast-gas composition.

Given. ZnS is oxidised per $\mathrm{2\,ZnS+3\,O_2\rightarrow2\,ZnO+2\,SO_2}$, with air supplied at 25 % excess over the stoichiometric oxygen requirement. Air is taken as 21 mol % O₂, 79 mol % N₂.

Find. The volume (mole) percent of SO₂, excess O₂ and N₂ in the gas leaving the roaster (ZnO reports to the solid calcine, not the gas).

Approach. Work on a basis of 2 mol ZnS, read the theoretical O₂ off the balanced equation, scale it by 1.25 for the actual air supplied, carry the associated N₂ through unreacted, and normalise every gas-phase species by their sum.

1. Basis and theoretical oxygen. On a basis of 2 mol ZnS, the balanced equation calls for $$n_{O_2}^{\text{theo}}=3\ \text{mol}$$
2. Scale to the actual air supplied. At 25 % excess, $$n_{O_2}^{\text{sup}}=3\times1.25=3.75\ \text{mol}$$ and the nitrogen that necessarily accompanies it, at the air ratio $79/21=3.762$, $$n_{N_2}=3.75\times\frac{79}{21}=14.107\ \text{mol}$$
3. Find the unreacted (excess) oxygen. Only the theoretical 3 mol is consumed by the reaction; the rest passes through: $$n_{O_2}^{\text{exc}}=3.75-3.00=0.75\ \text{mol}$$
4. Read the SO₂ produced directly off the stoichiometry. The equation gives 2 mol SO₂ per 2 mol ZnS reacted, so on this basis $$n_{SO_2}=2.00\ \text{mol}$$ (ZnO is a solid and takes no further part in the gas balance.)
5. Total the gas phase and normalise. $$n_{\text{gas}}=n_{SO_2}+n_{O_2}^{\text{exc}}+n_{N_2}=2.00+0.75+14.107=16.857\ \text{mol}$$ $$\boxed{y_{SO_2}=\frac{2.00}{16.857}\times100=11.86\ \%,\quad y_{O_2}=\frac{0.75}{16.857}\times100=4.45\ \%,\quad y_{N_2}=\frac{14.107}{16.857}\times100=83.69\ \%}$$ The three percentages sum to 100.00 %, confirming the normalisation.