Answer:
The percent yield for Br₂ in the reaction = 96.15%
Explanation:
The balanced stoichiometric equation for the reaction is given as
2 NaBr + 1 Cl₂ → 2 NaCl + 1 Br₂
Percent yield
= 100% × (Actual yield)/(Theoretical yield)
To find the theoretical yield,
5.29 g of NaBr reacts with excess chlorine
gas; this means that NaBr is the limiting reagent because it is used up in the process of the reaction, hence, it determines the amount of products to be found.
So, we convert the 5.29 g of NaBr into number of moles.
Number of moles = (Mass)/(Molar mass)
Molar Mass of NaBr = 102.894 g/mol
Number of moles = (5.29/102.894) = 0.0514121329 = 0.05141 mole
From the stoichiometric balance of the reaction,
2 moles of NaBr give 1 mole of Br₂
0.05141 mole of NaBr will give (0.05141×1/2) mole of Br₂; that is, 0.0257 mole of Br₂
Theoretical yield = Mass of Br₂ expected from the reaction
= (Number of moles) × (Molar mass)
Molar mass of Br₂ = 159.808 g/mol
Theoretical yield of Br₂ = 0.0257 × 159.808 = 4.108 g
Percent yield
= 100% × (Actual yield)/(Theoretical yield)
Actual yield = 3.95 g
Theoretical yield = 4.108 g
Percent yield = 100% × (3.95/4.108) = 96.15%
Hope this Helps!!!
