1 mole of Cl2 reacts with 2 moles of KBr to produce 1 mole of Br2.
The molar mass of Cl2 is 70.9 g/mol, and the molar mass of Br2 is 159.8 g/mol.
Therefore, 200 g of Cl2 is equal to 200 g / 70.9 g/mol = 2.82 moles of Cl2.
According to the stoichiometry, 2.82 moles of Cl2 will react with 2 x 2.82 = 5.64 moles of KBr to produce 2.82 moles of Br2.
The theoretical yield of Br2 is therefore 2.82 moles x 159.8 g/mol = 452.3 g.
Now, we can calculate the percentage yield:
Percentage yield = (Actual yield / Theoretical yield) x 100%
The actual yield is given as 410 g.
Percentage yield = (410 g / 452.3 g) x 100% = 90.6%
Therefore, the percentage yield of Br2 in this reaction is 90.6%.
