Question

What volume of hydrogen and oxygen measured at S.T.P. will be required to prepare 1.8 g of water?

Answer 1

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