Hydrogen sulphide (H2S) contains 94.11% sulphur,water (H2O) contains 11.11% hydrogen and sulphur dioxide (S02) contain 50% oxygen. Show that the results are in agreement with law of reciprocal proportions.
Answers
The weight of Hydrogen in water = 11.11 grams.
The weight of Oxygen in water = 100 - 11.11 = 88.89 g
The weight of Sulphur in Sulphur di oxide = 50 g
The weight of Oxygen in Sulphur di oxide = 100 - 50 = 50g
The weight of Sulphur that combines with 88.89g of Oxygen
= 50/50 * 88.89 = 88.89g
The ratio between the weights of sulphur and hydrogen which combine with a fixed weight of oxygen = 88.89 : 11.11 = 8:1 .......... (1)
The weight of Sulphur in hydrogen sulphide = 94.11g
The weight of Hydrogen = 100 - 94.11g = 5.89g
The ratio between the weights of sulphur and hydrogen
is 94.11 : 5.89 =16:1 ..................................(2)
From equations (1) and (2) we get,
8/1 : 16/1 = 8:16 = 1:2 ....( 3 )
From equation (3 ) we can come to inference that these are simple multiples of each other.
Therefore, it is in agreement with law of reciprocal proportions.
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