Let's work through the problem step by step, using **1 ton (100 kg) of copper concentrate** as the basis. --- # **a) Weight of Calcine per Ton of Concentrate** Let: - \( W_{conc} = 100 \) kg (basis) - \( W_{calcine} \) = unknown (to be found) **Elemental Balance Method:** Let's use a non-volatile element (e.g., **Fe**) for the calculation, since its total amount (except for oxidation state) doesn't change during roasting. #### **Fe in Concentrate:** \[ \text{Fe in concentrate} = 29.7\% \times 100 \text{ kg} = 297 \text{ kg} \] #### **Fe in Calcine:** \[ \text{Fe in calcine} = 37.6\% \times W_{calcine} \] Set equal: \[ 297 = .376 \times W_{calcine} \implies W_{calcine} = \frac{297}{.376} = 790.4 \text{ kg} \] --- # **b) Composition of the Calcine** Given in the problem (by wt.%): | Component | Wt.% | |-----------|------| | Cu | 11.1 | | SiO₂ | 24. | | Fe | 37.6 | | S | 10. | | Al₂O₃ | 6.7 | | CaO | .4 | | O₂ | 10.2 | --- # **c) % Sulphur Eliminated** #### **Sulphur in Concentrate:** \[ \text{S in concentrate} = 36.9\% \times 100 = 369 \text{ kg} \] #### **Sulphur in Calcine:** \[ \text{S in calcine} = 10.\% \times 790.4 = 79.04 \text{ kg} \] #### **Sulphur Eliminated:** \[ \text{S eliminated} = 369 - 79.04 = 289.96 \text{ kg} \] #### **% Sulphur Eliminated:** \[ \% \text{ eliminated} = \frac{289.96}{369} \times 100 = 78.6\% \] --- # **d) Volume and Composition of Gases (@STP)** ### **Mass of Sulphur Oxidized:** \[ \text{S oxidized} = 289.96 \text{ kg} \] #### **To SO₂:** (85%) \[ \text{To SO}_2 = .85 \times 289.96 = 246.47 \text{ kg} \] #### **To SO₃:** (15%) \[ \text{To SO}_3 = .15 \times 289.96 = 43.49 \text{ kg} \] --- ### **Moles of Gaseous Products** #### **SO₂:** \[ n_{SO_2} = \frac{246470}{64.07} = 3847.9 \text{ mol} \] #### **SO₃:** \[ n_{SO_3} = \frac{43490}{80.07} = 543.3 \text{ mol} \] --- ### **CO₂ from Fuel:** \[ \text{C burned} = 5 \text{ kg} = 500 \text{ g} \] \[ n_C = \frac{500}{12} = 416.7 \text{ mol} \] \[ \text{CO}_2 \text{ formed} = 416.7 \text{ mol} \] --- ### **N₂ from Air:** \[ \text{Air supplied} = 500 \text{ m}^3 \] At STP, 1 mol gas = 22.4 L = .0224 m³ \[ \text{Moles of air} = \frac{500}{.0224} = 223,214 \text{ mol} \] Air is 21% O₂, 79% N₂ by volume. #### **N₂:** \[ n_{N_2} = .79 \times 223,214 = 176,339 \text{ mol} \] --- ### **O₂ Consumed:** #### **From SO₂ production:** \[ \text{S} + \text{O}_2 \rightarrow \text{SO}_2 \] 246.47 kg S = 246,470 g / 32.07 = 7686.8 mol S\ O₂ used = 7686.8 mol #### **From SO₃ production:** \[ 2\text{S} + 3\text{O}_2 \rightarrow 2\text{SO}_3 \Rightarrow 1\text{S}:1.5\text{O}_2 \] 43.49 kg S = 43,490 g / 32.07 = 1356.1 mol S\ O₂ used = 1.5 × 1356.1 = 2034.15 mol #### **From Carbon oxidation (CO₂):** \[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \] O₂ used = 416.7 mol #### **Total O₂ consumed:** \[ \text{O}_2 \text{ used: } = 7686.8 + 2034.15 + 416.7 = 10,137.65 \text{ mol} \] #### **O₂ supplied:** \[ \text{O}_2 \text{ supplied: } .21 \times 223,214 = 46,875 \text{ mol} \] #### **O₂ remaining (in gases):** \[ \text{O}_2 \text{ remaining: } 46,875 - 10,137.65 = 36,737.35 \text{ mol} \] --- ### **Summary Table** | Gas | Moles | |-------|---------| | SO₂ | 3847.9 | | SO₃ | 543.3 | | CO₂ | 416.7 | | N₂ | 176,339 | | O₂ | 36,737 | #### **Total moles:** \[ \text{Total moles} = 3847.9 + 543.3 + 416.7 + 176,339 + 36,737 = 217,884.9 \text{ mol} \] #### **Volume at STP:** \[ \text{Total volume} = 217,884.9 \times .0224 = 4,877 \text{ m}^3 \] --- # **e) Excess Air Used** #### **O₂ Required:** - S oxidized: \( 10,137.65 \) mol O₂ - C oxidized: \( 416.7 \) mol O₂ \[ \text{Total O}_2 \text{ required} = 10,137.65 \text{ mol} \] \[ \text{O}_2 \text{ supplied} = 46,875 \text{ mol} \] #### **Excess O₂:** \[ \text{Excess O}_2 = 46,875 - 10,137.65 = 36,737.35 \text{ mol} \] #### **% Excess Air:** \[ \% \text{ excess air} = \frac{\text{O}_2 \text{ supplied} - \text{O}_2 \text{ required}}{\text{O}_2 \text{ required}} \times 100 = \frac{46,875 - 10,137.65}{10,137.65} \times 100 = 362\% \] --- # **Summary Table** | Part | Result | |------|--------| | (a) | Weight of calcine: **790.4 kg per 100 kg concentrate** | | (b) | Given in problem (see table above) | | (c) | % Sulphur eliminated: **78.6%** | | (d) | Volume and composition of gases: **4877 m³ at STP**<br>SO₂: 3848 mol, SO₃: 543 mol, CO₂: 417 mol, N₂: 176,339 mol, O₂: 36,737 mol | | (e) | **362% excess air** used | --- **If you need further breakdowns (e.g., per component calculation), let me know!**