Question

An industry is required to clean up 4.0 t of a by-product containing both toxic and inert material. The toxic content is 15%; the rest is inert. The by-product is treated with 55 t of solvent resulting in an amount of dirty solvent containing 0.35% toxic material and a discard containing 1.2% toxic material, some solvent, and all the inert materials. Determine the amount of dirty solvent produced, the percentage of solvent in the discard, and the mass of toxic substance removed in the discard.

Answer 1

Amount of dirty solvent produced: 40.9 t

Percentage of solvent in the discard: 83%

Mass of toxix substance in the discard: 0.20 t

Step-by-step explanation:

The mass of the by product plus the mass of the solvent must be equal to the mass of dirty solvent (Mds) plus the mass of the discard (Md) (global mass balance), so, in kg:

4,000 + 55,000 = Mds + Md

Mds + Md = 59,000 kg

Mds = 59,000 - Md (equation 1)

The mass of the toxic substance in the by-product must be the same of the sum of the mass of it in the dirty solvent and in the discard (toxic substance mass balance):

0.15x4,000 = 0.0035Mds + 0.012Md

0.0035Mds + 0.012Md = 600

Let's replace the equation 1:

0.0035x(59,000 - Md) + 0.012Md = 600

206.5 - 0.0035Md + 0.012Md = 600

0.023 Md = 393.5

Md = 17,109 kg

Md = 17.1 t of discard

Mds = 59,000 - 17,109

Mds = 41,891 kg = 40.9 t of dirty solvent

The solvent mass balance will be:

55 = (1 - 0.0035)Mds + S*Md, where S is the percentage of solvent is the discard, and the percentage of the solvent in the dirty solvent is 1 less the percentage of the toxic substance in it.

55 = 0.9965x40.9 + 17.1S

17.1S = 14.24

S = 0.83 = 83%

The mass of the toxic substance in the discard is:

0.012Md = 0.012x17.1= 0.20 t of toxic substance