CHEMN-Assume a sewage treatment plant processing 1

Assume a sewage treatment plant processing 1 million liters of wastewater per day containing 200 mg/L of degradable biomass, CH2O. Calculate the minimum volume of dry air at 25 degree that must be pumped into the wastewater per day to provide the oxygen required to

Solution

1

Calculate the mass of degradable biomass: First, we need to calculate the total mass of degradable biomass (CH2O) in the wastewater. The concentration is given as 200 mg/L, and the total volume is 1 million liters. The mass (m) can be calculated using the formula

m = C \times V

, where

C

is the concentration and

V

is the volume

2

Convert mass to moles: To find the moles of CH2O, we use the molar mass of CH2O, which is approximately 30 g/mol. The mass of CH2O in grams is converted to moles by dividing by the molar mass:

n = \frac{m}{M}

, where

n

is the number of moles and

M

is the molar mass

3

Stoichiometry of the reaction: The balanced chemical equation for the degradation of CH2O with oxygen (O2) is

CH2O + O2 \rightarrow CO2 + H2O

. From the equation, we see that 1 mole of CH2O requires 1 mole of O2 for complete combustion

4

Calculate moles of O2 required: Since the stoichiometry of the reaction is 1:1, the moles of O2 required will be equal to the moles of CH2O calculated in step 2

5

Convert moles of O2 to volume: To find the volume of dry air needed, we use the ideal gas law,

PV = nRT

, where

P

is the pressure,

V

is the volume,

n

is the number of moles,

R

is the ideal gas constant (0.0821 L·atm/mol·K), and

T

is the temperature in Kelvin. Assuming standard atmospheric pressure (1 atm), we can solve for

V

6

Adjust for the percentage of O2 in dry air: Dry air contains approximately 21% oxygen by volume. Therefore, the volume of air needed is the volume of O2 divided by 0.21

Answer

[Insert final answer here]

Key Concept

Stoichiometry and the ideal gas law are used to calculate the volume of air required to provide oxygen for the degradation of biomass.

Explanation

The calculation involves determining the mass of biomass, converting it to moles, using the stoichiometry of the reaction to find the moles of oxygen needed, and then using the ideal gas law to find the volume of oxygen at the given conditions, which is then adjusted for the oxygen content in air.

Assume a sewage treatment plant processing 1 million liters of wastewater per day containing 200 mg/L of degradable biomass, CH2O. Calculate the minimum volume of dry air at 25 degree that must be pumped into the wastewater per day to provide the oxygen required to degrade the biomass.

Solution

1

Calculate the mass of degradable biomass: To find the mass of degradable biomass (CH2O) in the wastewater, we use the formula

m = C \times V

, where

C

is the concentration (200 mg/L) and

V

is the volume (1 million liters)

2

Determine the stoichiometry of the reaction: The balanced chemical equation for the degradation of CH2O with oxygen is

CH2O + O2 \rightarrow CO2 + H2O

. From this, we see that 1 mole of CH2O requires 1 mole of O2 for complete degradation

3

Calculate the moles of CH2O: The molar mass of CH2O is 30.026 g/mol. Convert the mass of CH2O to moles using

n = \frac{m}{M}

, where

n

is the number of moles and

M

is the molar mass

4

Calculate the moles of O2 required: Since the stoichiometry is 1:1, the moles of O2 required will be equal to the moles of CH2O

5

Calculate the volume of O2 at 25°C and 1 atm: Use the ideal gas law

V = \frac{nRT}{P}

, where

R

is the ideal gas constant (0.0821 L·atm/mol·K),

T

is the temperature in Kelvin (298 K), and

P

is the pressure (1 atm)

6

Calculate the volume of dry air needed: Since air contains approximately 21% O2 by volume, divide the volume of O2 by 0.21 to find the total volume of air required

Answer

[Insert final answer here]

Key Concept

Stoichiometry and gas laws are essential for calculating the volume of air required to provide oxygen for the degradation of biomass.

Explanation

The calculation involves determining the mass of biomass, converting it to moles, using stoichiometry to find the moles of oxygen needed, and then using the ideal gas law to find the volume of oxygen at given conditions, which is then used to calculate the volume of air needed.

Assume a sewage treatment plant processing 1 million liters of wastewater per day containing 200 mg/L of degradable biomass, CH2O. Calculate the minimum volume of dry air at 25 degree that must be pumped into the wastewater per day to provide the oxygen required to degrade the biomass.

Solution

1

Calculate the mass of degradable biomass: To find the mass of degradable biomass in the wastewater, multiply the volume of wastewater by the concentration of biomass

$ \text{Mass of biomass} = \text{Volume of wastewater} \times \text{Concentration of biomass} = 1,000,000 \, \text{L} \times 200 \, \text{mg/L} $

2

Stoichiometry of the degradation reaction: The balanced chemical equation for the degradation of biomass (CH2O) with oxygen is:

\text{CH}_2\text{O} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}

. From this, we can see that 1 mole of CH2O requires 1 mole of O2 for complete degradation

3

Calculate moles of biomass: Convert the mass of biomass to moles using the molar mass of CH2O (30.03 g/mol)

$ \text{Moles of biomass} = \frac{\text{Mass of biomass}}{\text{Molar mass of CH2O}} = \frac{1,000,000 \times 200 \times 10^{-3}}{30.03} $

4

Calculate moles of oxygen needed: Since the stoichiometry is 1:1, the moles of oxygen needed are equal to the moles of biomass

$ \text{Moles of O}_2 = \text{Moles of biomass} $

5

Calculate volume of oxygen at STP: Use the ideal gas law to find the volume of oxygen at standard temperature and pressure (STP), where 1 mole of gas occupies 22.4 liters

$ \text{Volume of O}_2 \text{ at STP} = \text{Moles of O}_2 \times 22.4 \, \text{L/mol} $

6

Convert volume of oxygen to conditions at 25°C: Use the combined gas law to find the volume of oxygen at 25°C and 1 atm

$ \text{Volume of O}_2 \text{ at 25°C} = \text{Volume of O}_2 \text{ at STP} \times \frac{T_{25°C}}{T_{STP}} $

$ T_{25°C} = 298 \, \text{K} $ and $ T_{STP} = 273 \, \text{K} $

7

Calculate the volume of dry air needed: Since air is approximately 21% oxygen by volume, divide the volume of oxygen required by 0.21 to find the total volume of air needed

$ \text{Volume of air} = \frac{\text{Volume of O}_2 \text{ at 25°C}}{0.21} $

Answer

[Insert final answer here]

Key Concept

Stoichiometry and gas laws are essential for calculating the volume of air required to provide oxygen for the degradation of biomass in wastewater treatment.

Explanation

The calculation involves determining the mass of biomass, converting it to moles, using the stoichiometry of the degradation reaction to find the moles of oxygen needed, and then using the ideal and combined gas laws to calculate the volume of air at the specified conditions.

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