Question

What amount of ferrous sulphide wl be required to prepare 1.7 gm of h2s by reaction of ferrous sulphide with excess dilute

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

The amount of Ferrous sulfide required to prepare $1.7$ g of Hydrogen sulfide by the reaction of ferrous sulfide with excess dilute sulphuric acid is $4.4$ g·

Explanation:

Given that,

The mass of Hydrogen sulfide $=$ $1.7$ g

The balanced chemical equation for the given reaction is,

   $FeS\ +\ H_2SO_4\ --->\ FeSO_4\ +\ H_2S$

From the balanced chemical equation, it is clear that

One mole of Ferrous sulfide reacts with one mole of sulphuric acid gives one mole of Ferrous sulfate and one mole of Hydrogen sulfide·

We know,

One mole of Ferrous sulfide weigh $88$ g·

One mole of Hydrogen sulfide weigh $34$ g·

That is,

$88$ g of Ferrous sulfide gives $34$ g of Hydrogen sulfide·

Then,

To prepare $1.7$ g of Hydrogen sulfide we need $\frac{1.7\ *\ 88}{34}$ g of Ferrous sulfide·

That is,

⇒ $4.4$ g of Ferrous sulfide·

Therefore,

The amount of Ferrous sulfide required to prepare $1.7$ g of Hydrogen sulfide is $4.4$ g·