Let's break down the problem step by step using the data provided. --- ## **Given Data** **Ore concentrate (per 1 ton = 100 kg):** - Cu: 8.8% → 88 kg - SiO₂: 19% → 190 kg - Fe: 29.7% → 297 kg - S: 36.9% → 369 kg - Al₂O₃: 5.3% → 53 kg - CaO: .3% → 3 kg **Calcine composition (wt.%):** - Cu: 11.1% - SiO₂: 24% - Fe: 37.6% - S: 10% - Al₂O₃: 6.7% - CaO: .4% - O₂: 10.2% **Other conditions:** - 500 m³ air supplied per ton concentrate - 5 kg C per ton concentrate as fuel - Of S oxidized: 85% to SO₂, 15% to SO₃ --- ## **a) Weight of Calcine per Ton of Concentrate** Let \( x \) = weight of calcine (kg) The masses of Cu, SiO₂, Fe, Al₂O₃, CaO are conserved (assuming negligible volatilization). For example, Cu in concentrate = 88 kg Cu in calcine = 11.1% of \( x \) \[ .111x = 88 \implies x = \frac{88}{.111} \approx 793 \text{ kg} \] **Check with SiO₂:** \[ .24x = 190 \implies x = \frac{190}{.24} \approx 792 \text{ kg} \] **Check with Fe:** \[ .376x = 297 \implies x = \frac{297}{.376} \approx 790 \text{ kg} \] **Check with Al₂O₃:** \[ .067x = 53 \implies x = \frac{53}{.067} \approx 791 \text{ kg} \] **Check with CaO:** \[ .004x = 3 \implies x = \frac{3}{.004} = 750 \text{ kg} \] Most consistent values are ~790–793 kg. **Take average:** \[ x \approx 792 \text{ kg (rounded)} \] --- ## **b) Composition of the Calcine** Already given in the problem (by weight %): - Cu: 11.1% - SiO₂: 24% - Fe: 37.6% - S: 10% - Al₂O₃: 6.7% - CaO: .4% - O₂: 10.2% --- ## **c) % Sulphur Eliminated** Sulphur in concentrate: 369 kg Sulphur in calcine: \(.10 \times 792 = 79.2\) kg \[ \text{Sulphur eliminated} = 369 - 79.2 = 289.8 \text{ kg} \] \[ \% \text{ Sulphur eliminated} = \frac{289.8}{369} \times 100 \approx 78.6\% \] --- ## **d) Volume and Composition of Gases (at STP)** ### **Sulphur gases:** - S oxidized: 289.8 kg - 85% to SO₂: \(.85 \times 289.8 = 246.33\) kg S as SO₂ - 15% to SO₃: \(.15 \times 289.8 = 43.47\) kg S as SO₃ **Moles of SO₂:** \[ \text{Moles SO}_2 = \frac{246.33}{32} = 7.698\ \text{kmol S} \Rightarrow 7.698\ \text{kmol SO}_2 \] **Moles of SO₃:** \[ \text{Moles SO}_3 = \frac{43.47}{32} = 1.358\ \text{kmol S} \Rightarrow 1.358\ \text{kmol SO}_3 \] ### **CO₂ from fuel:** - Carbon burnt: 5 kg → all to CO₂ \[ \text{Moles C} = \frac{5}{12} = .417\ \text{kmol} \] \[ \text{CO}_2\ \text{formed} = .417\ \text{kmol} \] ### **O₂ consumption:** **For S oxidation:** \[ \text{S} + \text{O}_2 \rightarrow \text{SO}_2 \quad (7.698\ \text{kmol S}) \implies 7.698\ \text{kmol O}_2 \] \[ \text{S} + 1.5\text{O}_2 \rightarrow \text{SO}_3 \quad (1.358\ \text{kmol S}) \implies 2.037\ \text{kmol O}_2 \] **For C oxidation:** \[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \implies .417\ \text{kmol O}_2 \] **Total O₂ consumed:** \[ = 7.698 + 2.037 + .417 = 10.152\ \text{kmol O}_2 \] **O₂ in calcine:** \(10.2\% \times 792 \text{ kg} = 80.78 \text{ kg O}_2\) \[ \text{Moles O}_2\ \text{in calcine} = \frac{80.78}{32} = 2.524\ \text{kmol} \] ### **Air Supplied:** - 500 m³ air per ton concentrate - 1 mol ideal gas @STP = 22.4 L = .0224 m³ \[ \text{Moles air} = \frac{500}{.0224} = 223,214\ \text{mol} = 223.2\ \text{kmol} \] Air is ~21% O₂ by volume: \[ O_2\ \text{supplied} = .21 \times 223.2 = 46.87\ \text{kmol} \] --- **Gases Leaving:** - SO₂: 7.698 kmol - SO₃: 1.358 kmol - CO₂: .417 kmol - N₂ (from air): \(.79 \times 223.2 = 176.33\) kmol - O₂ unused: Supplied: 46.87 Consumed: 10.152 In calcine: 2.524 Remaining O₂ in gas: \(46.87 - 10.152 - 2.524 = 34.194\) kmol **Total gas moles:** \[ \text{Total} = 7.698 + 1.358 + .417 + 176.33 + 34.194 = 220.\ \text{kmol} \] \[ \text{Volume at STP} = 220. \times 22.4 = 4928\ \text{m}^3 \] --- ### **Summary Table of Gas Composition:** | Component | kmol | Volume % | |-----------|------|----------| | N₂ | 176.33 | 80.16% | | O₂ | 34.19 | 15.54% | | SO₂ | 7.70 | 3.50% | | SO₃ | 1.36 | .62% | | CO₂ | .42 | .19% | --- ## **e) Excess Air Calculation** **Theoretical O₂ required:** - For S: \(7.698 + 2.037 = 9.735\) kmol - For C: \(.417\) kmol - **Total theoretical O₂ needed:** \(9.735 + .417 = 10.152\) kmol **O₂ supplied:** 46.87 kmol \[ \text{Excess O}_2 = \frac{46.87 - 10.152}{10.152} \times 100 = 362\% \] **Excess air:** 362% --- ## **Summary of Answers** ### **a) Weight of calcine per ton of concentrate:** **792 kg** ### **b) Composition of calcine:** Given in the question. ### **c) % Sulphur eliminated:** **78.6%** ### **d) Volume and composition of gases (@STP):** **4928 m³ total** - N₂: 80.16% - O₂: 15.54% - SO₂: 3.5% - SO₃: .62% - CO₂: .19% ### **e) Excess air used:** **362% (excess O₂ supplied relative to theoretical requirement)** --- **If you need detailed step-by-step calculations, let me know!**