0.7 g of iron reacts directly with 0.4 g of sulphur to form ferrous sulphide. If 2.8 g of 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 ?

4:7

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6:4

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7:4

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8:4

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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 the same in both the cases, the law of constant composition is true.

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Similar questions

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 ?

What is the mass of sulphur that would react with 5 g of iron to form ferrous sulphide $(F e S)$ ?

What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, K_sp = 6.3 × 10^–18).

Q. What is the maximum concentration of an equimolar solution of ferrous sulphate

(F e S O_{4})

and sodium sulphide

(N a_{2} S)

so that when mixed in equal volumes, there is not precipitation of iron sulphide

(F e S)

?
(For FeS,

K_{s p} = 5 \times 10^{- 18}

)

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