Problem 1

In treating waste water in a sewage plant, 45 % of its carbohydrate,

(\ce{CH2O})_n

, is completely oxidized to

\ce{CO2}

and

\ce{H2O}

; 10 % undergoes anaerobic decomposition by fermentation into two gaseous components (

\ce{CO2}

and

\ce{CH4}

); the rest (45 %) remains in the sludge. The total gas formation is 16 m

^3

per day (at 25 °C, 100 kPa). 1.1 What is the amount of carbohydrate remaining in the sludge measured in kg per day? 1.2 Using the heat of combustion of methane (

-882

kJ mol

^{-1}

), calculate the amount of energy that can be produced by combustion of the methane formed per day. 1.3 Knowing that the concentration of the carbohydrate in the waste water is 250 mg dm

^{-3}

, calculate the daily amount of waste water processed in the plant in m

^3

of water per day.

Step 2 of 5: Gas-yielding reactions

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(\ce{CH2O})_n + n\,\ce{O2} \to n\,\ce{CO2} + n\,\ce{H2O};\quad (\ce{CH2O})_n \to \tfrac{n}{2}\,\ce{CO2} + \tfrac{n}{2}\,\ce{CH4}

Analysis

Complete oxidation converts each monomer unit $\ce{CH2O}$ into 1 mol of $\ce{CO2}$ gas (water is liquid); anaerobic fermentation converts each $\ce{CH2O}$ into $0.5\ \ce{CO2} + 0.5\ \ce{CH4} = 1$ mol of total gas. Thus every reacting monomer unit, whether oxidized (45 %) or fermented (10 %), yields exactly 1 mol of gas, so $45\,\% + 10\,\% = 55\,\%$ of the incoming carbohydrate accounts for the 646 mol of gas.