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?

45727g

Explanation:

So, have the overall ionic equation given as the following;

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

So, we have (from the question) that the amount or quantity of the waste stream daily = 60m^3/h, and the waste stream daily contains waste stream containing = 4.0 mg/L Cr, and the discharge limit = 0.1 mg/L.

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

Step two: Determine or calculate the the value of Cr used up.

The value of Car used up ={ 60,000 × ( 4.0 - 0.1) } ÷ 1000 = 234 g.

Step three: Determine or calculate the mass of Cr(OH)3 and the mass of Fe(OH)3.

The number of moles of Cr = 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.

Hence, the total = 463.5 g + 1441.8 g = 1905.3 g.

Step four: Determine or calculate the How much particulate matter would be generated daily.

The amount of the particulate that would be generated daily = 24 × 1905.3 = 45727g.