Question

what is the percent and theoretical yield if 3.50 g of NaBr and 6 Ml of 9 M H2SO4 ?

Answer 1

The balanced equation for the reaction taking place is:

2NaBr + 4H₂SO₄ ---> Br₂ + 2SO₂ + 4H₂O + 2NaSO₄

Moles of NaBr taken = Mass/MW = 3.5/102.89 = 0.034 moles

Moles of H2SO4 = Molarity*Volume = 9*0.006 = 0.054 moles

As per the reaction stoichiometry, sulfuric acid is the limiting reagent as it is in lesser amount than needed for complete reaction with NaBr.

So,

Theoretical Moles of SO₂ formed = 0.054/2 = 0.027 moles

Theroretical yield of SO₂ = Moles*MW = 0.027*64.06 = 1.73 g