The discharge of chromate ions (CrO42-) to sewers or natural waters is of concern because of both its ecological impacts and its

Question

2 months ago

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The discharge of chromate ions (CrO42-) to sewers or natural waters is of concern because of both its ecological impacts and its

effects on human health if the receiving water is later used as a drinking water source. One way in which chromate can be removed from solution is by its reaction with ferrous ions (Fe2+) to form a mixture of chromic hydroxide and ferric hydroxide solids [Cr(OH)3(s) and Fe(OH)3(s), respectively], which can then be filtered out of the water. The overall reaction can be represented as
CrO42- + 3 Fe2+ + 8 H2O --> Cr(OH)3(s) + 3 Fe(OH)3(s) + 4 H+
How much particulate matter would be generated daily by this process at a facility that treats 60 m3/h of a waste stream containing 4.0 mg/L Cr, if the treatment reduces the Cr concentration enough to meet a discharge limit of 0.1 mg/L?

Chemistry

1 answer:

KiRa [2.9K] · Answer 1 · 2026-03-10T22:20:09+03:00

Answer:

45727g

Explanation:

The overall ionic reaction can be described as follows:

CrO42^- + 3 Fe2^+ + 8 H2O ------> Cr(OH)^3(s) + 3 Fe(OH)^3(s) + 4 H^+.

The amount of wastewater processed each day is specified as 60m^3/h, with the concentration of chromium in the wastewater recorded at 4.0 mg/L, and the allowable discharge concentration is set at 0.1 mg/L.

Step one: Convert m^3/h to L/h. Therefore, 60 m^3/h × 1000 dm^3 = 60000 L/h.

Step two: Calculate the amount of chromium consumed.

The amount of chromium utilized = { 60,000 × ( 4.0 - 0.1) } ÷ 1000 = 234 g.

Step three: Calculate the masses of Cr(OH)3 and Fe(OH)3.

The moles of chromium = 234/52 = 4.5 moles.

Molar mass of Cr(OH)3 = 103 g/mol and the molar mass of Fe(OH)3 = 106.8 g/mol.

Thus, the mass of Cr(OH)3 = 4.5 × 103 = 463.5 g.

And, the mass of Fe(OH)3 = 13.5 × 106.8 = 1441.8 g.

Consequently, the total equals 463.5 g + 1441.8 g = 1905.3 g.

Step four: Calculate the amount of particulate matter generated each day.

The total particulate matter produced daily = 24 × 1905.3 = 45727g.