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Question
What volume of hydrogen and oxygen measured at S.T.P. will be required to prepare 1.8 g of water?
Solution
At S.T.P. \[\ce{2H2O -> 2H2 + O2}\]
2 vols. of water = 2 vols of Hydrogen
∴ 1 vol. of water = 1 vol. of Hydrogen
or, 18 gram of water = 22.4 lit.
∴ 1.8 gram of water = `22.4/18 xx 18/10`
= 2.24 lit. of H2
Again, 2 vols. of water = 1 vol. of oxygen
∴ 1 vol. of water = `1/2` vol. of O2
= `22.4/2`
= 11.2 lit.
∴ 18 gram of water = 11.2 lit. of O2
∴ 1.8 gram of water = `11.2/18 xx 18/10`
= 1.12 lit. of O2
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