The given chemical reaction is:

Δ
∑BE(reactants)-∑BE(products)
= {(941 kJ/mol) + (3 * 242 kJ/mol)} -[{2*(3*200 kJ/mol)}]
= 467 kJ/mol
Calculating the change in heat when 85.3 g chlorine reacts in the above reaction:
Moles of chlorine = 
= 1.20 mol 
Heat change when 1.20 mol chlorine reacts
= 
The sun heats the land
hot air rises
cold air from the ocean/body of water rushes in under the hot air
wind
Equations 2 and 4 are properly balanced
The reaction is:
Cl2 + 2 KBr --> 2 KCl + Br2
Moles of KCl is
n = m /M = 12 /74 = 0.16 mol
As, twice the moles of KCl is producing from 1 mol of chlorine
mole of Cl2 = 0.16 /2 = 0.08 mol
Mass of Cl2
m /70 = 0.08 = 5.6 g
Hence, 5.6 g mol Cl2 consumed to produce KCl