0.7 g of iron combine directly with 0.4 g of sulphur to form FeS . If 2.8g of Fe are dissolved in dil. HCl and excess of sodium sulphide sol is added , 4.4 g of FeS are precipitated . Show that this proves the law of constant proportion

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Solution

$M e t h o d I : F e + S \to F e S M a s s o f F e = 0.7 g M a s s o f S = 0.4 g 0.7 g o f F e r e a c t s w i t h 0.4 g o f S 1 g o f F e r e a c t s w i t h = \frac{0.4}{0.7} \times 1 = 0.5714 g o f S M e t h o d I I : M a s s o f F e = 2.8 g M a s s o f F e S = 4.4 g M a s s o f S = 4.4 - 2.8 = 1.6 g 2.8 g o f F e r e a c t s w i t h 1.6 g o f S 1 g o f F e r e a c t s w i t h = \frac{1.6}{2.8} \times 1 = 0.5714 g o f S A s b o t h m e t h o d s g i v e s s a m e a m o u n t o f S, 0.5714 g, l a w o f c o n s \tan t p r o p o r t i o n i s o b e y e d .$

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

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

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and sodium sulphide

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

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