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प्रश्न
LPG has 60% propane and 40% butane: 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to atmosphere.
\[\ce{C3H8 + 5O2 → 3CO2 + 4H2O}\]
\[\ce{2C4H10 + 13O2 → 8CO2 + 10H2O}\]
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उत्तर
Ratio of propane and Butane in mixture
= 60 : 40
= 3 : 2
Ratio in 10 litres mixture = `3/5 xx 10 and 2/5 xx 10`
= 6 lit : 4 lit
\[\ce{C3H8 + 5O2 → 3CO2 + 4H2O}\]
1 vol 3 vol
1 volume of propane produces CO2 = 3 volumes
∴ 6 litres produces CO2 = 3 x 6
= 18 litres ...(i)
\[\ce{2C4H10 + 13O2 → 8CO2 + 10H2O}\]
2 vol 8 vol
2 litres of Butane produces CO2 = 8 litres
∴ 4 litres produce CO2 = `(8 xx 4)/2`
= 16 litres ...(ii)
∴ CO2 added to atmosphere = 18 + 16
= 34 litres.
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संबंधित 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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What volume of oxygen would be required for complete combustion of 100L of ethane according to the following equation?
2C2H6 + 7O2 → 4CO2 + 6H2O
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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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\[\ce{2C2H2_{(g)} + 5O2_{(g)} → 4CO2_{(g)} + 2H2O_{(g)}}\]
The reaction: \[\ce{4N2O + CH4 -> CO2 + 2H2O + 4N2}\] takes place in the gaseous state. If all volumes are measured at the same temperature and pressure, calculate the volume of dinitrogen oxide (N2O) required to give 150 cm3 of steam.
1250 cc of oxygen was burnt with 300cc of ethane [C2H6]. Calculate the volume of unused oxygen formed:
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