Question

A coal sample consists of 82.1% carbon, 4.5% hydrogen, 1.5% sulphur, 3.0% oxygen and the remainder incombustible material. If 1 kg is burnt with 20% excess air, determine (i) the mass of air required per kilogram of fuel and (ii) prepare an analysis by mass of the products of combustion per kilogram of fuel.

Answer 1

Answer on Question #79764, Physics / Molecular Physics | Thermodynamics

A coal sample consists of 82.1 % carbon, 4.5 % hydrogen, 1.5 % sulphur, 3.0% oxygen and the remainder incombustible material. If 1 kg is burnt with 20 % excess air, determine (i) the mass of air required per kilogram of fuel and (ii) prepare an analysis by mass of the products of combustion per kilogram of fuel.

Solution

$$m(\text{fuel}) = 1\,\text{kg} = 1000\,\text{g}$$ $$w(C) = 0.821$$ $$w(H) = 0.045$$ $$w(S) = 0.015$$ $$w(O) = 0.03$$ $$w(O_2) \text{ in air} = 0.23$$ $$m_{\text{air}} - ?$$

Combustible substances in fuel are: carbon, hydrogen and sulphur. Complete oxidation of these elements leads to oxide formation.

(i) Determine the mass of air required per kilogram of fuel.

For carbon: $C + O_2 \rightarrow CO_2$

$$n(C) = \frac{m}{M} = \frac{w(C) \times m_{\text{air}}}{M} = \frac{0.821 \times 1000\,\text{g}}{12\,\frac{\text{g}}{\text{mol}}} = 68.4\,\text{mol}$$

The equation gives mole ratio $n(C) : n(O_2) = 1:1$, $n(O_2) = 68.4$ mol.

For sulphur: $S + O_2 \rightarrow SO_2$

$$n(S) = \frac{m}{M} = \frac{w(S) \times m_{\text{air}}}{M} = \frac{0.015 \times 1000\,\text{g}}{32\,\frac{\text{g}}{\text{mol}}} = 0.5\,\text{mol}$$

The equation gives mole ratio $n(S) : n(O_2) = 1:1$, $n(O_2) = 0.5$ mol.

For hydrogen: $4H + O_2 \rightarrow 2H_2O$

$$n(H) = \frac{m}{M} = \frac{w(H) \times m_{\text{air}}}{M} = \frac{0.045 \times 1000\,\text{g}}{1\,\frac{\text{g}}{\text{mol}}} = 45\,\text{mol}$$

The equation gives mole ratio $n(H) : n(O_2) = 4:1$, $n(O_2) = \frac{n(H)}{4} = 45\,\text{mol} / 4 = 11.3$ mol.

Total amount of substance of oxygen used for combustion of carbon, sulphur and hydrogen is:

$$\begin{array}{l} n_{\text{total}} = 68.4 \, \text{mol} + 0.5 \, \text{mol} + 11.3 \, \text{mol} = 80.2 \, \text{mol} \\ m(O_2) = n \times M = 80.2 \, \text{mol} \times 32 \, \text{g/mol} = 2566.4 \, \text{g} \\ w(O_2) = \frac{m(O_2)}{m_{\text{air}}} \Rightarrow m_{\text{air}} = \frac{m(O_2)}{w(O_2)} = \frac{2566.4 \, \text{g}}{0.23} = 11158.3 \, \text{g} \\ \end{array}$$

Excess of air is 20%, $m_{\text{excess iar}} = 11158.3 \, \text{g} \times 0.2 = 2231.7 \, \text{g}$

Total mass of air with 20% excess is $11158.3 \, \text{g} + 2231.7 \, \text{g} = 13390 \, \text{g} = 13.4 \, \text{kg}$

(ii) Prepare an analysis by mass of the products of combustion per kilogram of fuel.

Products of combustion are: CO₂, H₂O and SO₂

For CO₂:

$$\mathrm{C} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2$$

The equation gives mole ratio $n(\mathrm{C}) : n(\mathrm{CO}_2) = 1:1$, $n(\mathrm{CO}_2) = 68.4 \, \mathrm{mol}$.

$$m(\mathrm{CO}_2) = n \times M = 68.4 \, \text{mol} \times 44 \, \text{g/mol} = 3009.6 \, \text{g}$$

For SO₂:

$$\mathrm{S} + \mathrm{O}_2 \rightarrow \mathrm{SO}_2$$

The equation gives mole ratio $n(\mathrm{S}) : n(\mathrm{SO}_2) = 1:1$, $n(\mathrm{SO}_2) = 0.5 \, \mathrm{mol}$.

$$m(\mathrm{SO}_2) = n \times M = 0.5 \, \text{mol} \times 64 \, \text{g/mol} = 32 \, \text{g}$$

For H₂O:

$$4\mathrm{H} + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}$$

The equation gives mole ratio $n(\mathrm{H}) : n(\mathrm{H}_2\mathrm{O}) = 4:2$, $n(\mathrm{H}_2\mathrm{O}) = n(\mathrm{H})/2 = 45 \, \mathrm{mol}/2 = 22.5 \, \mathrm{mol}$.

$$m(\mathrm{H}_2\mathrm{O}) = n \times M = 22.5 \, \text{mol} \times 18 \, \text{g/mol} = 405 \, \text{g}$$

Combustion of 1 kg of fuel gives 3009.6 g of carbon dioxide, 405 g of water and 32 g of sulphur dioxide. The main product of combustion is carbon dioxide, the minor product is sulphur dioxide.

$$w(\mathrm{CO}_2) = \frac{3009.6}{3009.6 + 405 + 32} \times 100\% = 87.3\%.$$ $$w(\mathrm{SO}_2) = \frac{32}{3009.6 + 405 + 32} \times 100\% = 0.9\%.$$ $$w(\mathrm{H}_2\mathrm{O}) = \frac{405}{3009.6 + 405 + 32} \times 100\% = 11.8\%.$$

Answer: (i) 13.4 kg

(ii) Combustion of $1\mathrm{kg}$ of fuel gives $3009.6\mathrm{g}$ of carbon dioxide, $405\mathrm{g}$ of water and $32\mathrm{g}$ of sulphur dioxide. The main product of combustion is carbon dioxide, the minor product is sulphur dioxide.

$$w(CO_2) = 87.3\%, w(SO_2) = 0.9\%, w(H_2O) = 11.8\%.$$

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