For the reaction represented by the equation cl2 + 2kbr ® 2kcl + br2, calculate the percentage yield if 200. g of chlorine react with excess potassium bromide to produce 410. g of bromine.

The  %  yield if 200g   of chlorine  react  with  excess Potassium bromide to  produce  410g of  bromine  is calculated as below

%  yield =  actual  yield/theoretical yield  x100
the  actual yield = 410 grams

calculate the  theoretical yield
by first  calculate the moles of chlorine used

mole=  mass/molar  mass
molar  mass of Cl2 =  35.5 x2= 71 g/mol

moles=  200g/71g/mol  = 2.82  moles

cl2 +2 KBr = 2KCl  +Br2

by use  of mole ratio between Cl2  to Br2  which is 1:1  the  moles of  Br2 is also = 2.82  moles

theoretical  mass =  moles x molar mass
molar mass of Br2= 79.9 x2= 159.8 g/mol

moles= 2.82g x 159.8 g/mol = 450.64 grams

% yield is therefore = 410g/450.64 x100 = 90.98 %

Your question is not clear but if you can explain yourself i would be glad to .  just trying to : ))

B? i feel like this is a trick