The company you work for plans to release a waste stream containing 7 mg/L
of benzaldehyde . Assuming that benzaldehyde degrades according to the reaction:
C7H6O + 8O2 → 7CO2 + 3H2O
Calculate the theoretical carbonaceous oxygen demand of this waste stream in mg/L.
Molecular Weights:
C7H6O = 106g/mol = 106mg/mmol
O2 = 32g/mol = 32mg/mmol
Stoichiometry Demand:
8 moles of O2 are needed per 1 mole of C7H6O
Therefore:
(8mol-O2)/(1mol-C7H6O) * (32g/mol)/(106g/mol) * 7mg/L
All units will cancel except mg/L:
8*(32/106)*7 = 16.906 mg/L
Well, it seems like your company is planning to release a "waste stream" of benzaldehyde. It's like they're saying, "Hey, let's turn our waste into a stream of chemical equations!" Now, according to the reaction you provided, one molecule of benzaldehyde reacts with 8 molecules of oxygen to produce 7 molecules of carbon dioxide and 3 molecules of water.
So, let's do some math. Given that the concentration of benzaldehyde in the waste stream is 7 mg/L, we need to calculate the amount of benzaldehyde that can react with the oxygen.
To calculate the amount of benzaldehyde in the waste stream in moles/L, we divide the concentration in mg/L by the molar mass of benzaldehyde (106.12 g/mol).
7 mg/L ÷ 106.12 g/mol = 0.066 moles/L
Now, since the reaction coefficient of benzaldehyde is 1 in the balanced equation, we can say that 0.066 moles of benzaldehyde will react with 8 * 0.066 = 0.528 moles of oxygen.
Finally, to convert this to mg/L, we multiply the moles by the molar mass of oxygen (32 g/mol) and divide by the volume (in L).
0.528 moles/L * 32 g/mol = 16.896 g/L or 16,896 mg/L
So, the theoretical carbonaceous oxygen demand of the waste stream is around 16,896 mg/L. That's a big demand! I hope your company has a good plan to tackle that.
To calculate the theoretical carbonaceous oxygen demand (COD) of the waste stream, we need to determine the amount of benzaldehyde that will be converted to carbon dioxide (CO2) based on the given reaction.
The reaction indicates that 1 mole of benzaldehyde (C7H6O) requires 8 moles of oxygen (O2) to produce 7 moles of carbon dioxide (CO2) and 3 moles of water (H2O).
From the molecular formula of benzaldehyde (C7H6O), we can determine its molar mass as:
C = 12.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol
Molar mass of benzaldehyde (C7H6O) =
(7 x Molar mass of carbon) + (6 x Molar mass of hydrogen) + (1 x Molar mass of oxygen)
= (7 x 12.01 g/mol) + (6 x 1.01 g/mol) + (16.00 g/mol)
= 91.07 g/mol
To find the theoretical COD, we can calculate the amount of benzaldehyde (in moles) in the waste stream and then multiply it by the stoichiometric factor from the balanced reaction equation.
Given concentration of benzaldehyde in the waste stream = 7 mg/L
Convert mg/L to moles/L:
1 mole of benzaldehyde = 91.07 g
1 mg = 1/1000 g
7 mg = (7/1000) g
Concentration of benzaldehyde in moles/L =
(7/1000 g) / (91.07 g/mol)
Now, we can multiply the concentration of benzaldehyde (in moles/L) by the stoichiometric factor from the reaction to determine the theoretical COD.
The stoichiometric factor for benzaldehyde in the reaction is 7:
C7H6O + 8O2 → 7CO2 + 3H2O
Theoretical COD in moles/L =
Concentration of benzaldehyde in moles/L x Stoichiometric factor
= [(7/1000 g) / (91.07 g/mol)] x 7
Finally, we convert the theoretical COD from moles/L to mg/L by multiplying it by the molar mass of carbon (12.01 g/mol) to get the theoretical carbonaceous oxygen demand in mg/L.
Theoretical COD in mg/L =
Theoretical COD in moles/L x Molar mass of carbon (12.01 g/mol)
Performing the calculations will give you the value of the theoretical carbonaceous oxygen demand in mg/L for the waste stream.
To calculate the theoretical carbonaceous oxygen demand of the waste stream, we need to determine the number of moles of benzaldehyde in 1 liter of the waste stream and then multiply it by the stoichiometric coefficient from the given reaction.
1. Determine the number of moles of benzaldehyde in 1 liter of the waste stream:
The concentration of benzaldehyde is given as 7 mg/L. The molar mass of benzaldehyde (C7H6O) is:
(7 * 12.01 g/mol) + (6 * 1.01 g/mol) + (16.00 g/mol) = 106.12 g/mol
To convert milligrams to grams, divide by 1000:
7 mg/L = (7/1000) g/L = 0.007 g/L
Now, calculate the number of moles using the molar mass:
Number of moles = Mass / Molar mass = 0.007 g / 106.12 g/mol
2. Determine the carbonaceous oxygen demand based on the stoichiometry of the reaction:
From the balanced reaction:
1 molecule of benzaldehyde (C7H6O) requires 8 molecules of oxygen (O2) to produce 7 molecules of carbon dioxide (CO2) and 3 molecules of water (H2O).
The stoichiometric coefficient for benzaldehyde is 1, so the number of moles of benzaldehyde is multiplied by the stoichiometric coefficient ratio:
Carbonaceous oxygen demand = Number of moles of benzaldehyde * Stoichiometric coefficient ratio
In this case, the stoichiometric coefficient ratio is 8/1:
Carbonaceous oxygen demand = (0.007 g / 106.12 g/mol) * (8/1)
Finally, to convert from moles to milligrams, multiply the result by the molar mass of oxygen (O2):
Carbonaceous oxygen demand = (0.007 g / 106.12 g/mol) * (8/1) * (32.00 g/mol) = 0.148 mg/L
Therefore, the theoretical carbonaceous oxygen demand of the waste stream is 0.148 mg/L.