10 mL of `H_(2)` combine with 5 mL of `O_(2)` to form `H_(2)O` when 200 mL of `H_(2)` at S.T.P. is passed through heated CuO, latter loses 0.144 g of its weight. Does the above data correspond to the law of constant composition ?

Answer

Yes

Explanation

In the first experiment :
Ratio by volume of `H_(2) and O_(2)` combining to form `H_(2)O=2 :1`
In the second experiment :
`CuO+H_(2)overset(heat)(rarr)Cu+H_(2)O`
The loss weight of CuO is due to oxygen which has been removed. It is 0.44 g
Now, 32 g of `O_(2)` at S.T.P occupy = 22400 L
0.144 g of `O_(2)` at S.T.P occupy `= (("22400 mL"))/(("32 g"))xx(0.144g)=100.8mL`
Ratio by volume of `H_(2) and O_(2)` combining to form `H_(2)O = 200 : 100.8 = 2:1`
Since the two ratios are the same, the data corresponds to Law of Constant Composition.