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 #79808, 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

Find masses of carbon, hydrogen, sulphur and oxygen in 1 kg of a coal.

$$m(C) = 0.821 \times 1000 \, g = 821 \, g$$ $$m(H) = 0.045 \times 1000 \, g = 45 \, g$$ $$m(S) = 0.015 \times 1000 \, g = 15 \, g$$ $$m(O) = 0.030 \times 1000 \, g = 30 \, g$$

The reactions that take place in the process of burning:

$$C + O_2 \rightarrow CO_2$$ $$4H + O_2 \rightarrow 2H_2O$$ $$S + O_2 \rightarrow SO_2$$

From the equation (1) we can see that mole ratio is $n(C) : n(O_2) : n(CO_2) = 1 : 1 : 1$

Find amount of substance of carbon:

$$n = m / M$$ $$n(C) = \frac{821 \, g}{12 \, g/mol} = 68.4 \, mol$$

Then $n(O_2) = 68.4 \, mol$

$$m(O_2) = M(O_2) \times n(O_2) = 32 \, g/mol \times 68.4 \, mol = 2188.8 \, g$$ $$n(CO_2) = 68.4 \, mol$$ $$m(CO_2) = M(CO_2) \times n(CO_2) = 44 \, g/mol \times 68.4 \, mol = 3009.6 \, g$$

From the equation (2) we can see that mole ratio is $n(H) : n(O_2) : n(H_2O) = 4 : 1 : 2$,

then $n(O_2) = n(H)/4$, $n(H_2O) = n(H)/2$

Find $n(H) : n(H) = m(H) / M(H) = 45 \, g / 1 \, g/mol = 45 \, mol$

$$n(O_2) = 45 \, mol/4 = 11.3 \, mol$$ $$\begin{array}{l} m \left(O _ {2}\right) = M \left(O _ {2}\right) \times n \left(O _ {2}\right) = 3 2 g / m o l \times 1 1. 3 m o l = 3 6 1. 6 g \\ n \left(H _ {2} O\right) = 4 5 \mathrm {m o l} / 2 = 2 2. 5 \mathrm {m o l} \\ m \left(H _ {2} O\right) = M \left(H _ {2} O\right) \times n \left(H _ {2} O\right) = 1 8 g / m o l \times 2 2. 5 m o l = 4 0 5 g \\ \end{array}$$

From the equation (3) we can see that mole ratio is $n(S):n(O_2):n(SO_2) = 1:1:1$

Find n(S):

$$\begin{array}{l} n (S) = m (S) / M (S) = 1 5 g / 3 2 g / m o l = 0. 4 6 9 m o l \\ n \left(O _ {2}\right) = 0. 4 6 9 \mathrm {m o l} \\ m \left(O _ {2}\right) = 3 2 g / m o l \times 0. 4 6 9 m o l = 1 5. 0 g \\ n \left(S O _ {2}\right) = 0. 4 6 9 \mathrm {m o l} \\ m \left(S O _ {2}\right) = M \left(S O _ {2}\right) \times n \left(S O _ {2}\right) = 6 4 g / m o l \times 0. 4 6 9 m o l = 3 0. 0 g \\ \end{array}$$

So, determine (i) the mass of air required per kilogram of fuel:

Find the mass of oxygen required for burning $1\mathrm{kg}$ of coal

$$m \left(O _ {2}\right) = 2 1 8 8. 8 g + 3 6 1. 6 g + 1 5. 0 g = 2 5 6 5. 4 g$$

The oxygen content of air by mass is $23\%$ .

Then $m_{\text{air}} = 2565.4 \, \text{g} / 0.23 = 11153.9 \, \text{g}$

As air was with $20\%$ excess then $\mathsf{m}_{\mathrm{air}} = 11153.9 + 11153.9 \times 0.2 = 13384.7 \, \text{g} = 13.4 \, \text{kg}$

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

From the table data we can see that mass of $\mathrm{CO}_{2}$ formed is the bigger, the mass of $\mathrm{H}_{2} \mathrm{O}$ formed is $3009.6 / 405 = 7.4$ times less than the mass of $\mathrm{CO}_{2}$ , and the mass of $\mathrm{SO}_{2}$ is the minor (3009.6/30= 100 times less than the mass of $\mathrm{CO}_{2}$ )

Answer: (i) 13.4 kg

(ii)

From the table data we can see that mass of $\mathrm{CO}_{2}$ formed is the bigger, the mass of $\mathrm{H}_{2} \mathrm{O}$ formed is $3009.6 / 405 = 7.4$ times less than the mass of $\mathrm{CO}_{2}$ , and the mass of $\mathrm{SO}_{2}$ is the minor (3009.6/30= 100 times less than the mass of $\mathrm{CO}_{2}$ )

Answer provided by https://www.AssignmentExpert.com