Question

10 mL of H₂ combine with 5 mL of O₂ to form water. When 200 mL of H₂ 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?

Answer 1

In the second reaction, 0.144 g weight is lost from CuO which is due to reduction of CuO into Cu. We can say that 0.144 g oxygen is combined with 200 mL H₂.

\(\because\) 32 g oxygen occupies 22400 mL volume at STP

\(\therefore\) 0.144 g oxygen will occupy = \(\frac{2240}{32}\) x 0.144

= 100.8 mL of O₂

\(\therefore\) Ratio of H₂ and O₂ in water = 200 : 100.8

= 2 : 1

This ratio is equivalent to first reaction (i.e. 10 : 5 or 2 : 1).

Hence, the above data corresponds to the law of constant composition.