Mole ConceptHard
Question
A quantity of 5.08 g of iodine held in suspension in water is slowly acted upon by 460 ml of H2S measured at 0o C and 1 atm. What weight of sulphur will be liberated? (I = 127)
Options
A.0.64 g
B.0.657 g
C.1.297 g
D.0.017 g
Solution
$I_{2} + H_{2}S \rightarrow 2HI + S$
$\frac{5.08}{254} \frac{460}{22400} 0.02\text{ mole}$
= 0.02 mole = 0.0205 mole 0.02 × 32
(L.R.) = 0.64 g
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