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Agata [3.3K]
3 years ago
6

Match the procedural step to its purpose in the written procedure for the Fischer esterification of benzoic acid with methanol i

n the presence of mineral acid.
a. Flat-bottomed flasks transfer heat in an uneven fashion, and may fracture if heated strongly.
b. Methanol, which is used as a reagent and solvent, is flammable.
c. The unreacted carboxylic acid was neutralized under mild conditions.
d. Water needs to be removed before the liquid ester product is collected.

1. Magnesium sulfate is added to the dichloromethane layer before the solvent is distilled off.
2. The reagents were added to a round bottom flask prior to the reaction being heated under reflux.
3. The solution of benzoic acid and methanol was heated on a steam bath.
4. The reaction mix was washed with sodium carbonate solution.
Chemistry
1 answer:
Art [367]3 years ago
4 0

Answer:

a. 2. The reagents were added to a round bottom flask prior to the reaction being heated under reflux.

b. 3. The solution of benzoic acid and methanol was heated in a steam bath

c. 4. The reaction mix was washed with sodium carbonate solution.

d. 1. Magnesium sulfate is added to the dichloromethane layer before the solvent is distilled off.

Explanation:

Fisher esterification is an equilibrium process. There are unreacted carboxylic acids in the reaction mix which are weak acids and they neutralize into aqueous solution. Magnesium sulfate is drying agent there water should be removed from dichloromethane solution before the solvent is distilled off.

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Explanation:

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Molar mass of KIO_{3} = 214 g/mol

So, 562 g of KIO_{3} = \frac{562}{214} moles of KIO_{3} = 2.63 moles of KIO_{3}

Molarity of a solution = (number of moles of solute in solution)/(total volume of solution in liter)

Here solute is KIO_{3} and solvent is water

Total volume of solution is 4.30 L

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100 \frac{lb}{h}*\frac{453,5g}{1 lb}*\frac{1molH20}{18,016g}*\frac{1h}{3600s}=0,7\frac{molH20}{s}

ii) Now we just consider the oxygen in the water stream (for 1 mole of water there is 1 mole of oxygen):

100\frac{lb}{h} *\frac{453,5g}{1lb}*\frac{1 molH20}{18,016g}*\frac{1molO}{1molH20}*\frac{16gr}{1molO}*\frac{1h}{3600s}=11,2\frac{gO}{s}

iii)Just considering the hydrogen in the stream (for 1 mole of water there is 2 moles of hydrogen):

100\frac{lb}{h} *\frac{453,5g}{1lb}*\frac{1 molH20}{18,016g}*\frac{2molH}{1molH20}*\frac{1gr}{1molH}*\frac{1h}{3600s}=1,4\frac{gH}{s}

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