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At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL of air containing 20% O2 by volume for complete combustion. After combustion, the gases occupy 345 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is
- a)C3H6
- b)C3H8
- c)C4H8
- d)C4H10
Correct answer is option 'B'. Can you explain this answer?
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At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL of ...
Given Data:
Temperature (T) = 300 K
Pressure (P) = 1 atm
Volume of hydrocarbon (V_hydrocarbon) = 15 mL
Volume of air (V_air) = 375 mL
Volume after combustion (V_combustion) = 345 mL
Oxygen concentration in air = 20% by volume
Calculating the moles of O2:
The volume of air required for complete combustion is 375 mL, and the oxygen concentration in air is 20%. Therefore, the volume of oxygen (V_O2) in air is given by:
V_O2 = V_air × (20/100) = 375 mL × (20/100) = 75 mL
To calculate the moles of O2, we need to convert the volume to moles using the ideal gas equation:
n_O2 = (V_O2 × P) / (R × T)
where R is the ideal gas constant (0.0821 L·atm/(mol·K))
Substituting the given values:
n_O2 = (75 mL × 1 atm) / (0.0821 L·atm/(mol·K) × 300 K)
n_O2 ≈ 3.03 × 10^-3 mol
Calculating the moles of hydrocarbon:
We can use the ideal gas equation to calculate the moles of hydrocarbon (n_hydrocarbon) before combustion:
n_hydrocarbon = (V_hydrocarbon × P) / (R × T)
n_hydrocarbon = (15 mL × 1 atm) / (0.0821 L·atm/(mol·K) × 300 K)
n_hydrocarbon ≈ 6.07 × 10^-4 mol
Calculating the moles of water:
Since the hydrocarbon completely reacts with oxygen to form water, the moles of water (n_water) formed will be equal to the moles of hydrocarbon (n_hydrocarbon):
n_water = n_hydrocarbon ≈ 6.07 × 10^-4 mol
Calculating the moles of gases after combustion:
The volume after combustion is given as 345 mL, which is the sum of the volumes of water vapor and the remaining unreacted gases. Let's assume the volume of unreacted gases is V_unreacted.
V_combustion = V_water + V_unreacted
345 mL = (n_water × 22.4 L/mol) + (V_unreacted × P) / (R × T)
Substituting the values:
345 mL = (6.07 × 10^-4 mol × 22.4 L/mol) + (V_unreacted × 1 atm) / (0.0821 L·atm/(mol·K) × 300 K)
345 mL = 0.0136 L + (V_unreacted × 1) / (0.0821 × 300)
345 mL = 0.0136 L + 0.0017 V_unreacted
Simplifying the equation:
0.0017 V_unreacted = 345 mL - 0.0136 L
0.0017 V_unreacted = 0.3314 L
Temperature (T) = 300 K
Pressure (P) = 1 atm
Volume of hydrocarbon (V_hydrocarbon) = 15 mL
Volume of air (V_air) = 375 mL
Volume after combustion (V_combustion) = 345 mL
Oxygen concentration in air = 20% by volume
Calculating the moles of O2:
The volume of air required for complete combustion is 375 mL, and the oxygen concentration in air is 20%. Therefore, the volume of oxygen (V_O2) in air is given by:
V_O2 = V_air × (20/100) = 375 mL × (20/100) = 75 mL
To calculate the moles of O2, we need to convert the volume to moles using the ideal gas equation:
n_O2 = (V_O2 × P) / (R × T)
where R is the ideal gas constant (0.0821 L·atm/(mol·K))
Substituting the given values:
n_O2 = (75 mL × 1 atm) / (0.0821 L·atm/(mol·K) × 300 K)
n_O2 ≈ 3.03 × 10^-3 mol
Calculating the moles of hydrocarbon:
We can use the ideal gas equation to calculate the moles of hydrocarbon (n_hydrocarbon) before combustion:
n_hydrocarbon = (V_hydrocarbon × P) / (R × T)
n_hydrocarbon = (15 mL × 1 atm) / (0.0821 L·atm/(mol·K) × 300 K)
n_hydrocarbon ≈ 6.07 × 10^-4 mol
Calculating the moles of water:
Since the hydrocarbon completely reacts with oxygen to form water, the moles of water (n_water) formed will be equal to the moles of hydrocarbon (n_hydrocarbon):
n_water = n_hydrocarbon ≈ 6.07 × 10^-4 mol
Calculating the moles of gases after combustion:
The volume after combustion is given as 345 mL, which is the sum of the volumes of water vapor and the remaining unreacted gases. Let's assume the volume of unreacted gases is V_unreacted.
V_combustion = V_water + V_unreacted
345 mL = (n_water × 22.4 L/mol) + (V_unreacted × P) / (R × T)
Substituting the values:
345 mL = (6.07 × 10^-4 mol × 22.4 L/mol) + (V_unreacted × 1 atm) / (0.0821 L·atm/(mol·K) × 300 K)
345 mL = 0.0136 L + (V_unreacted × 1) / (0.0821 × 300)
345 mL = 0.0136 L + 0.0017 V_unreacted
Simplifying the equation:
0.0017 V_unreacted = 345 mL - 0.0136 L
0.0017 V_unreacted = 0.3314 L
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Community Answer
At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL of ...
Volume of CxHy = 15 mL
Final volume = Volume of unreacted air + volume of carbon dioxide formed.
Therefore, volume of carbon dioxide formed = 345 - 300 = 45 mL
According to Avogadro's law, volume ratio = molar ratio
Hence, x = 3
y = 8
Hence the hydrocarbon is propane (C3H8).
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At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL of air containing 20% O2by volume for complete combustion. After combustion, the gases occupy 345 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon isa)C3H6b)C3H8c)C4H8d)C4H10Correct answer is option 'B'. Can you explain this answer?
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