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muminat
1 year ago
8

the synthesis of aspirin is a common esterification reaction. it is based on the balanced reaction given in the background of th

is lab. calculate the theoretical yield of aspirin given that 3.52-ml of acetic anhydride was reacted with 0.9457-g of salicylic acid.
Chemistry
1 answer:
Alex Ar [27]1 year ago
6 0

The theoretical yield of aspirin is <u>1.23 g</u>

Aspirin is a common drug with many uses. It relieves pain and reduces the risk of serious problems such as heart attack and stroke. It comes in many forms, including:

It is important to remember that aspirin alone does not lower blood pressure. However, its ability to thin the blood may benefit some people with high blood pressure.

moles of salicylic acid = mass / molar mass

=> 0.9457 / 138.2 = <u>0.006843 mol</u>

Acetic anhydride is in excess. Thus salicylic acid is the limiting reactant.

Total moles of aspirin that can form = moles of salicylic acid = 0.006843

mass of aspirin = number of moles of aspirin x molar mass of aspirin

= 0.006843 mol x 180.16g/mol = 1.23g

Thus theoretical yield of aspirin = 1.23 g

Learn more about aspirin here:-brainly.com/question/13533428

#SPJ4

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2 years ago
PLS ANSWER ASAP The following reaction occurs in a car’s catalytic converter.
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3 years ago
What type of reaction is: CaO + H2O -&gt; Ca(OH)2? A. Combustion B. Synthesis C. Double Replacement D. Decomposition
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3 years ago
If 842 grams of sodium hydroxide reacts with 750.0 grams of aluminum, how many grams of aluminum hydroxide should theoretically
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548.55 grams of aluminum hydroxide should theoretically form.

Explanation:

Balanced equation for the reaction:

3 NaOH + Al ⇒ Al(OH)3 +3 Na

DATA GIVEN:

mass of NaOH = 842 grams, atomic mass =39.9 grams/mole

mass of Al = 750 grams, atomic mass = 26.9 grams/mole

aluminum hydroxide theoretical yield = ?

Moles of NaOH reacted

number of moles = \frac{mass}{atomic mass of 1 mole}

putting the values in the equation

NaOH = \frac{842}{39.9}

           = 21.1 MOLES OF NaOH

Al = \frac{750}{26.9}

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from the equation

 from 3 moles of NaOH 1 mole of Al(OH)3 is produced

21.1 moles of NaOH will react to give x moles of Al(OH)3

\frac{1}{3} = \frac{x}{21.1}

7.03 moles of Al(OH)3 is formed.

and

1 mole of Al(OH)3 is formed from 1 mole of Al in the reaction

so, 27.8 Moles will react to give give 27.8 moles of Al(OH)3 limiting reagent of the given reaction is NaOH

mass of Al(OH)3 =7.03 x 78 (atomic mass of Al(OH)3)

          = 548.55 grams

theoretical  yield from the given data is 548.55 grams

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