`10 mL` of hydrogen combine with `5mL` of oxygen to yield water. When 200 mL of hydrogen at N.T.P. are passed over heated `CuO`, the latter loses `0.144 g` of its mass. Do these results agree with the law of constant composition ?

Answer

A

Explanation

First method
First case
Weight of `10 mL` of `H_(2)` at `STP = (10 xx 2)/(22400) = 0.0008 g`
Weight of `5 mL` of `O_(2)` at `STP = (5 xx 32)/(22400) = 0.0007 g`
Second case
Weight of `200 m`L of `H_(2)` at `STP = (200 xx 2)/(22400) = 0.1178 g`
Weight of oxygen taken away from copic oxide by `200 mL` of `0.0178 g` of `H_(2)` at `STP = 0.144 g`
Weight of `O_(2)` that combines with `0.0008 g` of `H_(2)` in first case `= (0.144 xx 0.0008)/(0.0178) = 0.0065 ~~ 0.007 g`
Thus, the weights of `O_(2)` that combine with same weight of `H_(2)` is the same in two cases. Hence, the law of constant composition is proved.
Second method
First case
`underset({:(1 mL),(10 mL),(200 mL):})(H_(2)) + underset({:((1)/(2) mL),(5 mL),(100 mL):})((1)/2) O_(2) rarr H_(2) O`
Second case
Weight of oxygen taken away from cupric oxide by `200 mL` of `H_(2)` at `STP = 0.144 g`
`:.` volume of `0.144 g of O_(2) = (22400 xx 0.144)/(32)`
`= 100.8 mL`
So the volume of `O_(2)` that combines with `H_(2)` in both the cases is same. Hence, the law of constant composition is verifed.