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At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 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 air...
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At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air...
Given Information:
- Volume of gaseous hydrocarbon = 15 mL
- Volume of air containing 20% O2 = 375 mL
- Volume of gases after combustion = 330 mL
Calculations:
1. Determine the volume of O2 in the air:
Volume of O2 = 375 mL * 20% = 75 mL
2. Volume of O2 consumed in combustion:
Volume of O2 consumed = Volume of initial gases - Volume of final gases
Volume of O2 consumed = (375 mL - 75 mL) - 330 mL = 270 mL
3. Volume of O2 consumed in combustion is equal to the volume of O2 required for the complete combustion of the hydrocarbon.
4. Determine the moles of O2 consumed:
n(O2) = Volume of O2 consumed / 22.4 L (1 mole of gas at STP)
n(O2) = 270 mL / 22.4 L = 0.012 mol
5. Since the hydrocarbon is completely combusted, it will produce CO2 and H2O.
The balanced equation for the combustion of hydrocarbon CnHm is:
CnHm + (n+m/4)O2 → nCO2 + m/2 H2O
6. Calculate the moles of hydrocarbon:
From the volume ratio of hydrocarbon to O2, we have:
n(hydrocarbon) / n(O2) = Volume of hydrocarbon / Volume of O2 consumed
n(hydrocarbon) = (15 mL / 270 mL) * 0.012 mol = 0.00067 mol
7. From the moles of hydrocarbon obtained, it corresponds to the formula C3H8. Hence, the formula of the hydrocarbon is C3H8.
- Volume of gaseous hydrocarbon = 15 mL
- Volume of air containing 20% O2 = 375 mL
- Volume of gases after combustion = 330 mL
Calculations:
1. Determine the volume of O2 in the air:
Volume of O2 = 375 mL * 20% = 75 mL
2. Volume of O2 consumed in combustion:
Volume of O2 consumed = Volume of initial gases - Volume of final gases
Volume of O2 consumed = (375 mL - 75 mL) - 330 mL = 270 mL
3. Volume of O2 consumed in combustion is equal to the volume of O2 required for the complete combustion of the hydrocarbon.
4. Determine the moles of O2 consumed:
n(O2) = Volume of O2 consumed / 22.4 L (1 mole of gas at STP)
n(O2) = 270 mL / 22.4 L = 0.012 mol
5. Since the hydrocarbon is completely combusted, it will produce CO2 and H2O.
The balanced equation for the combustion of hydrocarbon CnHm is:
CnHm + (n+m/4)O2 → nCO2 + m/2 H2O
6. Calculate the moles of hydrocarbon:
From the volume ratio of hydrocarbon to O2, we have:
n(hydrocarbon) / n(O2) = Volume of hydrocarbon / Volume of O2 consumed
n(hydrocarbon) = (15 mL / 270 mL) * 0.012 mol = 0.00067 mol
7. From the moles of hydrocarbon obtained, it corresponds to the formula C3H8. Hence, the formula of the hydrocarbon is C3H8.
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At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 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)C3H6b)C3H8c)C4H8d)C4H10Correct answer is option 'B'. Can you explain this answer?
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