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112 mL of hydrogen combines with 56 mL o...

`112 mL` of hydrogen combines with `56 mL` of oxygen of form water. When `224 mL` of hydrogen is passes over hand cupric oxide, the cupric oxide loses. `0.160 g` of weight. All volumes are measured at `STP`. Show that the result agrees with the law of constant composition `(22.4 L` hydrogen and oxygen at `STP` weigh, respectively, `2g` and `32 g`)

Text Solution

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Weigth of `112 mL` of `H_(2)` at `STP = (112 xx 2)/(22400) = 0.01 g`
Weight of `56 mL` of `O_(2)` at `STP = (56 xx 32)/(22400) = 0.08 g`
Second case :
Weight of `224 mL` of `H_(2)` at `STP = (224 xx 2)/(22400) = 0.02 g`
Weight of oxygen taken away from cupric oxide by `224 mL` or `0.02 g` of `H_(2)` at `STP = 0.160 g`
Weight of oxygen that combines with `0.01 g` of `H_(2)` in the second case `= (0.160 xx 0.01)/(0.02) = 0.08 g`
Thus, the two weigths of oxygen that combine with the same weight of hydrogen are same in the two cases. Hence, the law of constant composition is proved.
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