The question is complete, the complete question is;
16.1g of bromine are mixed with 8.42g of chlorite to give an actual yield of 21.1g of bromine monochloride.
Answer:
91 %
Step-by-step explanation:
Br2 + Cl2 ------>2BrCl
Number of moles of Chlorine = 8.42/71 = 0.119 moles
If 1 mole of Cl2 yields 2 moles of BrCl
0.119 moles of Cl2 will yield 0.119 moles * 2/1 = 0.238 moles of BrCl
Number of moles of Br2 reacted = 16.1/160 = 0.1 moles
If 1 mole of Br2 yielded 2 moles of BrCl
0.1 moles of Br2 yields 0.1 * 2/1 = 0.2 moles of BrCl
Bromine is the limiting reactant.
Mass of BrCl produced = 0.2 moles * 115.357 g/mol = 23.07 g
Theoretical yield = 23.07 g
Actual yield = 21.1 g
Percentage yield = Actual yield/Theoretical yield * 100
= 21.1/23.07 * 100 = 91 %