CH20 Waste
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Assume that a waste contains 395.0 mg/L of biodegradable CH2O and is processed through a 190000.0 L/day sewage treatment plant which converts 40% of the waste to CO2 and H2O. Calculate the volume(L) of air (at 25 degrees, 1 atm) required for this conversion. Assume that the O2 is transferred to the water with 20% efficiency.
Hint: Calculate mass of carbohydrate processed, use chemical equations to calculate mol oxygen needed to oxidize the carbohydrate, and the volume of air needed.
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Calculate the mass of CH2O or carbohydrate processed. The waste contains 395.0 mg/L but only 40% is converted in the process so 158 mg/L is converted. 191,000 L/day run through the plant so the total mass converted per day is 158 x 190,000 = 3.002x10^7 mg carbohydrate processed per day.
The chemical reaction is CH2O + O2 ------------>CO2 + H2O
so we need to convert the mass of carbohydrate to moles of cabohydrate ...
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