Let's break down the **copper ore roasting problem** step-by-step. For clarity, **all calculations are based on 1 ton (100 kg) of concentrate**. --- ## 1. **Given Data** ### **Concentrate (per ton, 100 kg)** | Component | wt.% | Mass (kg) | |-----------|------|-----------| | Cu | 8.8 | 88 | | SiO₂ | 19 | 190 | | Fe | 29.7 | 297 | | S | 36.9 | 369 | | Al₂O₃ | 5.3 | 53 | | CaO | .3 | 3 | | **Total** | 100 | 100 | ### **Calcine (Unknown mass, let it be x kg)** | Component | wt.% | |-----------|------| | Cu | 11.1 | | SiO₂ | 24 | | Fe | 37.6 | | S | 10 | | Al₂O₃ | 6.7 | | CaO | .4 | | O₂ | 10.2 | | **Total** | 100 | --- ## (a) **Weight of Calcine per Ton of Concentrate** Let the mass of calcine = \( x \) kg **Mass balance for a non-volatile element (Al₂O₃ or SiO₂):** Let's use **Al₂O₃** (assuming no loss): \[ \text{Al}_2\text{O}_3 \text{ in concentrate} = \text{Al}_2\text{O}_3 \text{ in calcine} \] \[ 53 \text{ kg} = x \times \frac{6.7}{100} \] \[ x = \frac{53 \times 100}{6.7} = \underline{791} \text{ kg (rounded)} \] --- ## (b) **Composition of the Calcine** Already given as wt.% in the question. For mass (kg) in calcine, multiply wt.% by 791 kg: | Component | wt.% | Mass (kg) | |-----------|------|-----------| | Cu | 11.1 | 87.8 | | SiO₂ | 24 | 190 | | Fe | 37.6 | 297.4 | | S | 10 | 79.1 | | Al₂O₃ | 6.7 | 53 | | CaO | .4 | 3.2 | | O₂ | 10.2 | 80.6 | --- ## (c) **% Sulphur Eliminated** \[ \text{S in concentrate} = 369 \text{ kg} \] \[ \text{S in calcine} = 79.1 \text{ kg} \] \[ \text{S eliminated} = 369 - 79.1 = 289.9 \text{ kg} \] \[ \%\text{S eliminated} = \frac{289.9}{369} \times 100 = \underline{78.6\%} \] --- ## (d) **Volume and Composition of Gases (@STP)** ### **Sulphur to SO₂ and SO₃** - S eliminated: 289.9 kg = 9,048.8 mol - 1 mol S → 1 mol SO₂ or 1 mol SO₃ - **85% to SO₂**: \( .85 \times 9048.8 = 7691.5 \) mol - **15% to SO₃**: \( .15 \times 9048.8 = 1357.3 \) mol #### **SO₂:** \[ 7691.5 \text{ mol} \times 64.07\text{ g/mol} = 492,746 \text{ g} = 492.7 \text{ kg} \] #### **SO₃:** \[ 1357.3 \text{ mol} \times 80.07\text{ g/mol} = 108,682 \text{ g} = 108.7 \text{ kg} \] --- ### **Combustion of Carbon (Fuel)** \[ 5 \text{ kg C} \rightarrow \text{CO}_2 \] \[ \text{Moles C} = \frac{500}{12} = 416.7\text{ mol} \] \[ 416.7\text{ mol} \times 44.01 = 18,334\text{ g} = 18.3\text{ kg CO}_2 \] --- ### **O₂ Consumed** #### **For S:** \[ \text{For SO}_2: S + O_2 \rightarrow SO_2 \implies 7691.5 \text{ mol S} \rightarrow 7691.5 \text{ mol O}_2 \] \[ \text{For SO}_3: S + 1.5 O_2 \rightarrow SO_3 \implies 1357.3 \text{ mol S} \rightarrow 2035.95 \text{ mol O}_2 \] \[ \text{Total O}_2 \text{ for S} = 7691.5 + 2035.95 = 9727.45 \text{ mol} \] #### **For C:** \[ \text{C + O}_2 \rightarrow \text{CO}_2 \implies 416.7 \text{ mol O}_2 \] --- ### **Total O₂ Consumed:** \[ 9727.45 + 416.7 = 10,144.2 \text{ mol} = 324.6 \text{ kg} \] --- ### **Air Supplied** \[ 500 \text{ m}^3 \text{ air per ton} \] \[ \text{1 mol gas at STP} = 22.4 \text{ L} = .0224 \text{ m}^3 \] \[ 500 \div .0224 = 223,214 \text{ mol air} \] - Air: 21% O₂, 79% N₂ (by mol) \[ \text{O}_2 \text{ supplied} = 223,214 \times .21 = 46,875 \text{ mol} \] \[ \text{N}_2 \text{ supplied} = 223,214 \times .79 = 176,339 \text{ mol} \] --- ### **O₂ Remaining** O₂ in calcine: 80.6 kg = 2,518.75 mol O₂ in gas (unused): \[ \text{O}_2 \text{ supplied} - \text{O}_2 \text{ consumed} - \text{O}_2 \text{ in calcine} \] \[ = 46,875 - 10,144.2 - 2,518.75 = 34,212 \text{ mol} \] --- ### **Total Gas Volumes (at STP):** - SO₂: 7,691.5 mol - SO₃: 1,357.3 mol - CO₂: 416.7 mol - O₂ (unused): 34,212 mol - N₂: 176,339 mol **Total moles of gas:** \[ 7,691.5 + 1,357.3 + 416.7 + 34,212 + 176,339 = 220,016.5 \text{ mol} \] **Total volume at STP:** \[ 220,016.5 \times .0224 = \underline{4,928 \, \text{m}^3} \] --- **Composition by mole %:** | Component | Moles | Mole % | |-----------|-------|--------| | SO₂ | 7,691.5 | 3.5% | | SO₃ | 1,357.3 | .6% | | CO₂ | 416.7 | .2% | | O₂ | 34,212 | 15.6% | | N₂ | 176,339 | 80.1% | --- ## (e) **Excess Air Used** \[ \text{Theoretical O}_2 \text{ needed} = \text{O}_2 \text{ consumed by S and C} = 10,144.2 \text{ mol} \] \[ \text{Air needed theoretically} = \frac{10,144.2}{.21} = 48,306 \text{ mol} \] \[ \text{Air supplied} = 223,214 \text{ mol} \] \[ \text{Excess air} = \frac{223,214 - 48,306}{48,306} \times 100 = \underline{362\%} \] --- ## **Summary Table** | Item | Value | |----------------|------------------------------| | (a) Calcine | 791 kg per ton concentrate | | (b) Composition| Given in calcine table above | | (c) S removed | 78.6% | | (d) Gas volume | 4,928 m³ @ STP, see above | | (e) Excess air | 362% | --- ### **Let me know if you need any step in more detail or want to see a specific calculation expanded.**