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Question
0.7 g of iron reacts directly with 0.4 g of sulphur to form ferrous sulphide. If 2.8 g f iron dissolves in dilute HCl and excess of sodium sulphide solution is added, 4.4 g of iron sulphide is precipitated. If it obeys the law of constant composition, what is the ratio of weights of Fe : S ?
0.7 g of iron reacts directly with 0.4 g of sulphur to form ferrous sulphide. If 2.8 g f iron dissolves in dilute HCl and excess of sodium sulphide solution is added, 4.4 g of iron sulphide is precipitated. If it obeys the law of constant composition, what is the ratio of weights of Fe : S ?
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Solution
The correct option is C 7:4
The ratio of the weights of iron and sulphur in the first sample of the compound is Fe : S : : 0.7 : 0.4 or 7 : 4. According to the second experiment, 2.8 g of iron gives 4.4 g ferrous sulphide, or 2.8 g Fe combines with S = 4.4 - 2.8
= 1.6 g
Therefore, the ratio of the weights of Fe : S : : 2.8 : 1.6 or 7:4.
Since the ratio of the weight of the two elements is same in both the cases, the law of constant composition is true.
7:4
The ratio of the weights of iron and sulphur in the first sample of the compound is Fe : S : : 0.7 : 0.4 or 7 : 4. According to the second experiment, 2.8 g of iron gives 4.4 g ferrous sulphide, or 2.8 g Fe combines with S = 4.4 - 2.8
= 1.6 g
Therefore, the ratio of the weights of Fe : S : : 2.8 : 1.6 or 7:4.
Since the ratio of the weight of the two elements is same in both the cases, the law of constant composition is true.
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