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प्रश्न
24 cc Marsh gas (CH4) was mixed with 106 cc oxygen and then exploded. On cooling the volume of the mixture became 82 cc, of which, 58 cc was unchanged oxygen. Which law does this experiment support? Explain with calculations.
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उत्तर
This experiment supports Gay Lussac’s law of combining volumes.
According to Gay lussac’s law, the volumes of gases reacting should be in a simple ratio.
\[\ce{CH4 + 2O2 → CO2 + 2H2O}\]
1V 2V 1V 0V
24 cc 48 cc 24 cc
1 vol. of CH4 requires O2 = 2 vols.
∴ 24 cc of CH4 requires O2 = 2 × 24 = 48 cc
Similarly, 24 cc of CH4 produce CO2 = 24 cc
∴ Vol. of O2 unused = 106 − 48 = 58cc
∴ Vol. of mixture on cooling = unused oxygen + CO2 formed
= 58 + 24
= 82 cc
संबंधित प्रश्न
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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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\[\ce{2C2H2_{(g)} + 5O2_{(g)}-> 4CO2_{(g)} + 2H2O_{(g)}}\]
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\[\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}\]
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\[\ce{C3H8 + 5O2 → 3CO2 + 4H2O}\]
\[\ce{2C4H10 + 13O2 → 8CO2 + 10H2O}\]
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112 cm3 of H2S(g) is mixed with 120 cm3 of Cl2(g) at STP to produce HCl(g) and sulphur(s). Calculate the volume of gaseous product formed.
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