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A vessel of 10L was filled with 6 mole o...

A vessel of `10L` was filled with `6` mole of `Sb_(2)S_(3)` and `6` mole of `H_(2)` to attain the equilibrium at `440^(@)C` as:
`Sb_(2)S_(3)(s)+3H_(2)(g)hArr2Sb(s)+3H_(2)S(g)`
After equilibrium the `H_(2)S` formed was analysed by dissolving it in water and treating with excess of `Pb^(2+)` to give `708 g "of" PbS` as precipitate. What is value of `K_(c)` of the reaction at `440^(@)C`?(At. weight of `Pb=206)`.

A

`0.08`

B

`0.8`

C

`0.4`

D

`0.04`

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
Mole of `PbS=708//236="mole of" H_(2)S`
`Sb_(2)S_(3)(s)+3H_(2)(g)hArr2Sb(s)+3H_(2)S(g)`
`{:("Initial",6,6,0,0),("at eq.",5,3,2,3):}`
`K_(c)=((3//10)^(3)xx(2//10)^(2))/((5//10)xx(3//10)^(3))=(4)/(50)=0.08`
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