Show that the results given below taken together illustrate a law of chemical action: (a) `0.46 g` og amgnesium produces `0.77 g`. Of magnesium oxide, (b) `0.82 g` of magnesium liberates `760 mL` of hydrogen at `STP` from an acid (weight of `1 mL` of hydrogen at `STP = 0.00009 g)`, and (c ) `1.25 g` results from the union of `1.11 g` of oxygen and hydrogen.

Explanation

a. Magnesium oxide contains
`Mg = 0.46 g. O = 0.77 - 0.46 = 0.31 g`
Therefore, weight of oxygen that combines with `1 g` of
`Mg = (0.31)/(0.46) = 0.674 g`
b. `0.82 g` of `Mg` liberates `H_(2) = 0.00009 xx 760 = 0.068 g`
Thus, the weight of `H_(2)` that will be liberated by `1 g` of
`Mg = (0.068)/(0.82) = 0.083 g`
According to (a) and (b) the ratio of the weight of `O : H` that combine with or is displaced by the same weight of magnesium is `0.674 : 0.083` or `8 : 1`.
Again according (c ), `1.25 g` of water contains `O = 1.11 g` and `H = 1.25 - 1.11 = 0.14 g`
Therefore, The ratio of `O : H :: 1.11 : 0.14` or `8 : 1`
Hence, that ratio being the same, the law of reciprocal propertions is illustrated.