Hydrogen sulphide (H_(2)S) contains 94.11 % sulphur. Sulphur dioxide (SO_(2)) contains 50 % oxygen and water (H_(2)O) contains 11.11 % hydrogen. Shpw that the results are in agreement with the law of reciprocal proportions.

Hydrogen sulphide (H_(2)S) contains 94.11 % sulphur. Sulphur dioxide (

1.	Hydrogen sulphide (H_(2)S) contains 94.11 % sulphur. Sulphur dioxide (SO_(2)) contains 50 % oxygen and water (H_(2)O) contains 11.11 % hydrogen. Shpw that the results are in agreement with the law of reciprocal proportions.
Answer» Solution :Let US fix `1g` of sulphur (S) as the fixed WEIGHT In hydrogen sulphide `(H_(2)S)` Weight of sulphur `= 94.11 g` Weight of hydrogen `=100 - 94.11 =5.89 g` `94.11g` of sulphur have combined have compound with hydrogen `= 5.89 g` `1.0 g` sulphur has combined with hydrogen `= (5.89)/(94.11)g` In sulphur dioxide `(SO_(2))` `50.0 g` of sulphur have combined with OXYGEN `= 50.0 g` `1.0 g` of sulphur has combined with oxygen `= 1.0 g` Thus, the ratios by weight of hydrogen and oxygen combining with a fixed weight of sulphur in the two compounds is or `(5.89)/(94.11):1or5.89:94:11` (This ratio is not the same) In water `(H_(2)O)` Weight of hydrogen `= 11.11 g` Weight of oxygen `= 100-11.11 = 88.89 g` `11.11 : 88.89` Let us compare the ratios (i) and (II). There are related to each other as : `(5.89)/(94.11):(11.11)/(88.89)or0.0625:0.1336or1:2` Since these ratios are the simple whole number multiples of each other, the LAW of Reciprocal Proportion is illustrated.