Question

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0.7g of iron combine directly with 0.4 g of sulphur to form FeS.If 2.8 g of Fe are dissolved in dil HCL and excess of Na2S solution are added.4.4g of solution is precipitated.


. Prove that these data illustrate the law of constant composition Sodium sulphide.


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Answer 1

Explanation:

0.7 g of Fe combine    ⇒     0.4 g of S

To provide FeS ,

Mass of FeS  =  0.7 + 0.4

                     =  1.1 g

Now percentage of Fe in FeS

$\displaystyle{\% \ Fe=\dfrac{0.7}{1.1}\times100 }\\\\\\\displaystyle{\% \ Fe=63.63\%}$

In second case ;

2.8 g of Fe dissolve in HCl  and

4.4 g of FeS is as precipitated .

Now  percentage of Fe in FeS

$\displaystyle{\% \ Fe=\dfrac{2.8}{4.4}\times100 }\\\\\\\displaystyle{\% \ Fe=63.63\%}$

Thus the percentage of Fe is same in both case .

So according to given data it holds law of constant composition .

Hence proved .

Answer 2

Answer:

0.7 gram of iron combine directly with 0.4 gram of sulphur to form FeS .

The Mass of FeS

= 0.7 + 0.4

= 1.1 gram

The percentage :

Fe = 0.7 × 100

1.1

Fe = 63.63 %

2.8 gram of are dissolved in Dil HCL and excess of an Na2s solution are added 4.4 gram of solution precipitate :

Fe = 2.8 × 100

4.4

Fe = 63.63 %.

So, the percentage of Fe is same in all the two.

So, that data illustrate the law of constant composition Sodium sulphide .

So ,It is proved .