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 %