Question 2 of 6: Roasting a Pyrite / Zinc-Sulfide Charge — Cinder Balance

Question A2: Roasting a Pyrite / Zinc-Sulfide Charge — Cinder Balance (25 points)

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.

Given. Basis: 100 kg of mixed charge. “The pyrites yield 92% FeS₂” is read as the pyrite ore assaying 92% FeS₂. The cinder (solid residue) contains Fe₂O₃, ZnO, the gangue and inerts, SO₃ held in the solid, and some unoxidized FeS₂. Its sulfur is 3.5% of the cinder mass, with 70% of that sulfur present as SO₃ and 30% as FeS₂. Molar masses used: Fe 55.845, S 32.06, O 15.999, Zn 65.38.

Find. (a) the cinder mass and composition per 100 kg charge; (b) the sulfur remaining in the cinder as a percentage of the sulfur charged.

Approach. Let $C$ be the cinder mass. The 3.5% sulfur specification fixes the SO₃ and the unreacted FeS₂ as fractions of $C$. The FeS₂ that does react all ends up as Fe₂O₃, whichever of the first two reactions it follows. Summing the cinder species then gives one linear equation in $C$.

1. Zinc. The cinder analysis reports no residual ZnS, so all ZnS is roasted: $$m_{ZnO}=17.00\times\frac{81.379}{97.44}=14.20\ \text{kg}.$$
2. Sulfur species tied to $C$. The sulfur in the cinder is $0.035C$. Of this, $0.70(0.035C)=0.0245C$ is present as SO₃ and $0.0105C$ as FeS₂: $$m_{SO_3}=0.0245C\times\frac{80.057}{32.06}=0.06118\,C,\qquad m_{FeS_2,unr}=0.0105C\times\frac{119.965}{2(32.06)}=0.01964\,C.$$
3. Iron oxide. Two moles of FeS₂ give one mole of Fe₂O₃, i.e. $0.5(159.687)/119.965=0.6656$ kg Fe₂O₃ per kg FeS₂ oxidized: $$m_{Fe_2O_3}=0.6656\,(69.00-0.01964\,C)=45.92-0.01307\,C.$$
4. Cinder closure. $$C=m_{Fe_2O_3}+m_{ZnO}+m_{inerts}+m_{SO_3}+m_{FeS_2}=(45.92-0.01307C)+14.20+14.00+0.06118C+0.01964C,$$ so $C(1-0.06775)=74.12$ and $$\boxed{C=79.51\ \text{kg of cinder per 100 kg charge}}.$$
5. Composition. Back-substituting: FeS₂ $=0.01964(79.51)=1.56$ kg; SO₃ $=4.86$ kg; Fe₂O₃ $=0.6656(69.00-1.56)=44.88$ kg; ZnO 14.20 kg; gangue + inerts 14.00 kg. The masses sum to 79.51 kg. As a check, the sulfur in the SO₃ (1.948 kg) plus the sulfur in the FeS₂ (0.835 kg) is 2.783 kg, which is 3.50% of the cinder and splits 70/30 as specified.
6. (b) Sulfur left in the cinder. Sulfur charged: $$S_{FeS_2}=69.00\times\frac{64.12}{119.965}=36.88\ \text{kg},\qquad S_{ZnS}=17.00\times\frac{32.06}{97.44}=5.59\ \text{kg},\qquad S_{tot}=42.47\ \text{kg}.$$ Sulfur in the cinder $=0.035(79.51)=2.78$ kg, so $$\frac{2.78}{42.47}\times100=\boxed{6.55\%\ \text{of the sulfur charged stays in the cinder}}.$$ The other 93.45% (39.69 kg S) leaves in the burner gas as SO₂.

Most of the cinder is iron oxide. The inerts and the zinc oxide together make up about a third of it, and the sulfur-bearing species are only 8%. That is consistent with an efficient roast in which about 93% of the sulfur is released as SO₂ for acid-making.