In: Chemistry
From an initial mix of: 2 mol Cl2, 1 mol Br2, 1 mol I2, 1 mol Cl−, 1 mol Br−, & 5 mol I−; the total quantity of Br2 present after all possible reactions occur is
0 mol |
1 mol |
2 mol |
0.5 mol |
The order of oxidising strength or the tandency of different halogens to reduce is
Cl2 > Br2 > I2
Hence Cl2 oxidises both Br- and I- to Br2 and I2 respectively and itself reduced to Cl-.
Similarly, Br2 oxidises only I- to I2 and itself reduced to Br-.
and I2 cannot oxidise either of Cl- and Br-.
Hence the following ractions occur
Cl2 having highest oxidising strength, and I- having the highest reducing strength, they will first react with eachother.
Cl2 + 2I- ---> I2 + 2Cl-
Initial: 2mol 5mol 1 mol 1 mol
final: 0mol 1 mol 3 mol 5 mol
Now Cl2 shuld react with Br2, but since no Cl2 is left , it will not react with Br2.
Now Br2 having the second highest oxidising strength, and I- having the highest reducing strength, they will react with each other.
Br2 + 2I- ---> I2 + 2Br-
Initial: 1mol 1mol 3 mol 1 mol
final: 0.5mol 0 mol 3.5 mol 2 mol
In the above reaction, only 0.5 mol of B r2 will react, because there is only 1 mole of I- left from previous reaction and from the stoichiometry data, 1 mole of Br2 reacts with 2 moles of I-.
Hence total quantity of Br2 present after all possible reactions = 0.5 mol
Hence last option should be correct