Vikas TU
Last Activity: 7 Years ago
Dear Student,
200 ml of 0.2 M KI arrangement has 0.2/5= 0.04 mol KI
Here the adjusted condition is
2KI+ Cl2 = 2KCl + I2
In this way, for 2 moles KI there is need of 1 mole Cl2
For 0.04 mole it is 0.02 mole
Volume is 0.02x22.4 lit = 0.448 lit (STP).
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)