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^{+} +
BrO_{3}^{-} yielding ½
Br_{2} + 3 H_{2}O

( Br
^{-} yielding ½
Br_{2} + 2 e^{-} )
5 .

** 6 H ^{+} + 5 Br ^{-} +
BrO_{3}^{-} yielding 3
Br_{2} + 3 H_{2}O.**

**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, KBrO_{3} is the
limiting reagent in the Bromine generation.

From 1 mol of KBrO_{3} 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**.

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