Chemistry: A solution of carbon tetrachloride and carbon disulfide containing 50% by weight of each is to be continuously distilled at a rate of 5000 kg/h. (a) Determine the concentration of the mixture in terms of mole fractions. (b) Determine the average molecular weight of the mixture. (c) Calculate the feed rate in kmol/h.

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A solution of carbon tetrachloride and carbon disulfide

Ответ:

Chente379

22.01.2022

Chemistry

(a) CS₂ = 0.331, CCl₄ = 0.669; (b) 102 g/mol; (c) 49.1 kmol/h

Explanation:

Assume 5000 kg of mixture.

(a) Mole fractions

(i) Mass of each compound

Mass of CCl₄ = 2500 kg

Mass of CS₂ = 2500 kg

(ii) Moles of each compound

$\text{Moles of CCl}_{4} = \text{2500 kg CCl}_{4} \times \dfrac{\text{1 kmol CCl}_{4}}{\text{153.82 kg CCl}_{4}} = \text{16.3 kmol CCl}_{4}\\\\\text{Moles of CS}_{2} = \text{2500 kg CS}_{2} \times \dfrac{\text{1 kmol CS}_{2}}{\text{76.14 kg CS}_{2}} = \text{32.8 kmol CS}_{2}$

(iii) Mole fractions

Total moles = 16.3 kmol + 32.8 kmol = 49.1 kmol

The major component is usually called the solvent — Component 1 —carbon disulfide.

$\chi_{2} = \dfrac{n_{2}}{n_{\text{tot}}} =\dfrac{\text{16.3}}{\text{49.1}} = \mathbf{0.331}\\\\\chi_{1} = \dfrac{n_{1}}{n_{\text{tot}}} = \dfrac{\text{35.8}}{\text{49.1}} = \mathbf{0.669}$

The average molar mass is the weighted molar mass of each component.

$\text{MM} =\chi_{2} \text{M}_{2} + \chi_{1} \text{M}_{1} = 0.331 \times \dfrac{\text{153.82 g }}{\text{1 mol}} + 0.669 \times \dfrac{\text{76.14 g}}{\text{1 mol}} = \dfrac{\text{50.9 g}}{\text{1 mol }} + \dfrac{\text{50.9 g}}{\text{1 mol }}\\\\=\textbf{102 g/mol}$

(d) Feed rate

$\text{Feed rate} = \dfrac{\text{total moles}}{\text{time}} = \textbf{49.1 kmol/h}$

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