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प्रश्न
Propane burns in air according to the following equation:
C3H8 + 5O2 → 3CO2 + 4H2O
What volume of propane is consumed on using 1000 cm3 of air, considering only 20% of air contains oxygen?
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उत्तर
Given:
C3H8 + 5O2 → 3CO2 + 4H2O
Volume of air = 1000 cm3
Percentage of oxygen in air = 20%
From the given information
C3H8 + 5O2 → 3CO2 + 4H2O
1vol 5vols 3vols 4vols
According to Gay-Lussac’s law,
1 vol. of propane consumes 5 vol. of oxygen
Volume of oxygen = 1000 cm3 × 20% = 200 cm3
Therefore,
Volume of propane burnt for every 200 cm3 of oxygen
`= 1/5 xx 200 = 40 cm^3`
40 cm3 of propane is burnt.
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संबंधित प्रश्न
Propane burns in air according to the following equation:
\[\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}\]
What volume of propane is consumed on using 1000 cm3 of air, considering only 20% of air contains oxygen?
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\[\ce{2NO + O2 → 2NO2}\]
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\[\ce{4NH3 + 5O2 → 4NO + 6H2O}\]
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- If the volume of A is actually 5.6 dm3 at STP, calculate the number of molecules in the actual Volume of D at STP (Avogadro's number is 6 × 1023).
- Using your answer from (iv), state the mass of D if the gas is dinitrogen oxide (N2O).
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