10 mL of ${ H }_{ 2 }$ combine with 5 mL of
${ O }_{ 2 }$ to form water. When 200 mL of
${ H }_{ 2 }$ at STP is passed over
heated CuO, the CuO loses 0.144 g of its weight. Does the above data correspond to the law of constant composition?
In the second experiment 0.144 g weight is lost from CuO. This is due to the reduction of CuO into Cu. In other words, 0.144 g oxygen combined with 200 mL H 2.
32 g oxygen occupies 22400 mL volume at STP.
0.144 g oxygen will occupy = 22400 x 0.144/ 32 = 100.8 mL${{O}{2}}$
3
It means the ratio of H2 and ${{O}{2}}$ in water is 200 : 100.8 =2: 1. The same ratio is in first case (10:5 or 2:1). Thus, the law of constant composition is proved.