From an initial mix of: 2 mol Cl2, 1 mol Br2, 0.5 mol I2, 0.5 mol Cl−, 1 mol Br−, & 2 mol I−; the total quantity of Br− present after all possible reactions occur is:
| A)1.5 mol |
| B) | 1 mol |
| C) | 0.5 mol |
| D) | 0 mol |
Cl2 reacts with Br-
Cl2 + 2Br- --> Br2 + 2Cl-
1 mole of chlorine requires 2 moles of Br- ( but there is only 1 mole of Br-)
Hence this requires 0.5 moles of Chlorine which is availabel. Hence, all the Br- is consumed.
1 mole of Br- is consumed ( limiting reactant) and 0.5 mol of Cl2 is consued
and Br2 reacts with I2
Br2 + 2I2 --> I2 + 2Br- (2)
I2 is the limiting reactant and Br2 is excess
0.5 mole of I2 produces 0.5 moles of Br-
Br- present = 0.5 mol
C is the correct aswer
From an initial mix of: 2 mol Cl2, 1 mol Br2, 0.5 mol I2, 0.5 mol...
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