Question

# what is the theoretical yield of 1.35 g acetanilide with 6 mL acetic acid + 0.280...

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?

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.

#### Earn Coins

Coins can be redeemed for fabulous gifts.