Answer:
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.
