Question

The volume of chlorine at STP required to liberate all the bromine and iodine in 100 ml of 0.1 M each of KI and MBr₂ will be:
(a) 0.224 L
(b) 0.336 L
(c) 0.448 L
(d) 0.560 L

Answer 1

2KI + Cl2  ------->  2KCl + I2

Moles of KI reacted = 0.2 x 200/1000 = 0.04 moles of KI

According to reaction stoichiometry, 0.04moles of KI will react with 0.04/2 moles Cl2  to liberate I2 completely.

At STP, 1 mole Cl2 corresponds to 22.4L

So, 0.02 moles of Cl2  corresponds to 0.448L

 

Hence 0.448L of Cl2 will be required to liberate I2 completely from the given solution



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Answer 2

Answer:

2KI + Cl2  ------->  2KCl + I2

Moles of KI reacted = 0.2 x 200/1000 = 0.04 moles of KI

According to reaction stoichiometry, 0.04moles of KI will react with 0.04/2 moles Cl2  to liberate I2 completely.

At STP, 1 mole Cl2 corresponds to 22.4L

So, 0.02 moles of Cl2  corresponds to 0.448L

 

Hence 0.448L of Cl2 will be required to liberate I2 completely from the given solution