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 ?

10 mL of H_(2) combine with 5 mL of O_(2) to form H_(2)O when 200 mL o

1.	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» Solution :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.