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.
