Question

Imagine that three hundred pounds of dry sewage is dumped into a small lake, the volume of which is 300 million liters. How many tonnes of oxygen are needed to completely degrade this sewage? You may assume the sewage has an elemental composition of C6H12O6.

Answer 1

As the sewage has elemental composition of C6H12O6, so its degradation can be represented by the equation

C6H12O6 +6O2→6CO2+6H2O

moles of sewage=300 pounds=300 pounds*453.592 grams/pound=136077.6 grams

moles of sewage=136077.6 grams/molar mass

molar mass of C6H12O6=6*12+12*1+6*16=72+12+96=180 grams/mol

moles of sewage=136077.6 grams/molar mass=136077.6 grams/(180g/mol)=755.98 moles

From the balanced equation, you can figure out that 1 mol of sewage(C6H12O6) needs 6 moles of oxygen for degradation

So , 755.98 moles of sewage would require=6*755.98 moles=4535.88 moles of O2

Mass of O2 required=4535.88 moles*molar mass of O2=4535.88*16g/mol=72574.08 grams

Mass of O2 required (in tones)=72574.08*10^-6 tonnes=0.0725 tonnes of O2 [1gram=1*10^-6 tonnes]

Answer=0.0725 tonnes of oxygen