Show that the following experimental data are in agreement with the law of reciprocal proportions: (1) 0.46g of Mg on burning in air forms 0.76g of MgO.
(2) 0.41 g of Mg on reaction with excess of acid produces 380 `cm^(3)` of `H_(2)` at STP.

Explanation

(1) Mass of oxygen in 0.76g of `MgO=(0.76-0.46)=0.30`g
`therefore`Mass of oxygen combining with 1g of `Mg=(0.30)/(0.46)`
=0.652g
(2) Mass of 22400`cm^(3)` of `H_(2)` gas (at STP)=2g
`therefore`Mass of 380 `cm^(3)` of `H_(2)` gas (at STP)`=(2xx380)/(22400)=0.034g`
`therefre`Amount of `H_(2)` gas produced by 0.41g of Mg=0.034g
`therefore`Amoun of `H_(2)` gas produced by 1g of Mg
`=(0.034)/(0.41)=0.083g`
From (1) and (2), it is found that the masses of hydrogen and oxygen combined with or replaced by a fixed mass of Mg are in the ratio, 0.083:0.652`~~1.8` Now, according to the law off reciprocal proportions, if the elements H and O combine together, the ratio of their masses in the resulting compound will be either 1:8 or any simple multiple of it.