Question

what is the theoretical yield of 1.35 g acetanilide with 6 mL acetic acid + 0.280 g potassium bromate + 1 mL hydrobromic acid? molecular bromine is created in situ from potassium bromate and hydrobromic acid. I have the redox reaction completed:

5 e^- + 6 H⁺ + BrO₃^-     yielding       ½ Br₂ + 3 H₂O

                ( Br ^-       yielding             ½ Br₂ + 2 e^- ) 5       .

    6 H⁺ + 5 Br ^- + BrO₃^-    yielding         3 Br₂ + 3 H₂O.

​With my limiting reagent numbers, I get potassium bromate as the limiting reagent.. but then what? what is the ratio?

Answer 1

Potassium bromate: 0.280 g x mol/(167.0 g)=0.0017 mol

Hydrobromic acid: 1.0 mL x (1.49 g)/mL x mol/(80.912 g)=0.018 mol

Now, from the balanced chemical equation 1 mol Potassium Bromate reacts with 5 moles of HBr.

Hence, KBrO₃ is the limiting reagent in the Bromine generation.

From 1 mol of KBrO₃ under acidic condition 3 moles of Bromine are liberated.

Hence, available Bromine for Bromination is (3 * 0.0017) mol = 0.0051 mol

Now, during bromination of acetanilide 1 mol of bromine reacts with 1 mol of acetanilide.

Acetanilide used in this recation is 1.35 g x mol/(135.163 g)=0.0099 mol.

Therefore, Bromine is the limiting reagent during bromination.

Hence, we will get 0.0051 mol of 4-Bromo Acetanilide.

Molar mass of 4-Bromo Acetanilide 214.06 g/mol.

Therefore, mass of 4-Bromo Acetanilide theoritically produced is (0.0051 mol * 214.06 g/mol) = 1.09 g.