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Sagot :
Answer:
18.368 Liters
Explanation:
Given that balanced Acetone oxidation,
= C3H6O + 4O2 = 3CO2 + 3H2O
lets assume volume of acetone = 1 liter
mass of acetone = 50 mg
number of moles present = (5*10^-3)g / ( 58.08 ) = 0.0000861 mol
Given that : I mol of ( acetone ) = 4 mol of oxygen
∴0.0000861 mol of acetone = 0.0003444 mol of oxygen
also note that the ratio of air to oxygen is
0.21 mol of oxygen = 1 mol of air
∴ 0.0003444 mol of oxygen = 0.00164 mol of air (0.0003444 / 0.21 )
note : 1 mol at STP = 22.4 L (at STP = Temperature = O°C
Finally the theoretically oxygen demand in liters of air for a 50 mg/L of acetone to decompose
Given that Acetone decompose/degrade completely between 500°C - 600°
we will take 500° for the sake of this calculation
The oxygen demand in liters will be
= 22.4 * 0.00164 * 500 = 18.368 Liters
The theoretical oxygen demand in liters of air for a 50 mg / L acetone solution, to decompose completely is 18.368 Liters.
How we calculate moles?
Moles of any substance will be calculated as:
n = W/M, where
W = given mass
M = molar mass
Given chemical reaction is:
CH₃COCH₃ + 4O₂ → 3CO₂ + 3H₂O
Given mass of acetone = 50mg = 0.05g
Moles of acetone = 0.05g / 58.08g/mol = 0.0000861 mole
From the stoichiometry of the reaction, it is clear that:
1 mole of CH₃COCH₃ = 4 moles of O₂
0.0000861 mole of CH₃COCH₃ = 4×0.0000861=0.0003444 moles of O₂
We know that ratio of air is:
0.21 mole of oxygen present = in 1 mole of air
0.0003444 mole of oxygen present = in 0.0003444/0.21=0.00164 mole of air
At STP 1 mole of gas is present in 22.4 Liter.
Degradation of acetone takes place in 500°C - 600°C.
Theoretical oxygen demand = 22.4×0.00164×500 = 18.368 Liters
Hence, theoretical oxygen demand is 18.368 Liters.
To know more about theoretical oxygen demand, visit the below link:
https://brainly.com/question/13251445
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