The density of ethanoic anhydride is 1.092 g/mL. Use this value, and the amount that you used in the laboratory to show that ethanoic anhydride is in excess for your preparation of ASA?, 2.5ml of ethanoic anhydride was used.
To determine if ethanoic anhydride is in excess for the preparation of Acetylsalicylic acid (ASA), we need to calculate the mass of the ethanoic anhydride used and compare it to the theoretical amount required.
1. Calculate the mass of ethanoic anhydride used:
Density = mass/volume
Rearrange the equation to solve for mass:
Mass = density x volume
Mass = 1.092 g/mL x 2.5 mL
Mass = 2.73 grams
2. Since the molar mass of ethanoic anhydride (C4H6O3) is 102.09 g/mol, we can calculate the theoretical amount required:
Theoretical amount = molar mass x moles
Moles = mass / molar mass
Moles = 2.73 g / 102.09 g/mol
Moles = 0.0267 mol
The stoichiometry of the reaction between ethanoic anhydride and salicylic acid is 1:1, meaning that 1 mole of ethanoic anhydride reacts with 1 mole of salicylic acid to produce ASA.
3. Assuming that salicylic acid is the limiting reagent, we can use the balanced chemical equation to find the molar amount of ASA formed:
C4H6O3 + C7H6O3 -> C9H8O4 + CH3COOH
1 mol + 1 mol -> 1 mol + 1 mol
Moles of ASA formed = 0.0267 mol
4. Finally, we compare the amount of ethanoic anhydride used to the amount required for the reaction:
Since the mole ratio between ethanoic anhydride and ASA is 1:1, the molar amount of ASA formed should ideally be equal to the molar amount of ethanoic anhydride used.
Moles of ethanoic anhydride used = 0.0267 mol
Moles of ASA formed = 0.0267 mol
Since the moles are equal, we can conclude that ethanoic anhydride is in excess for the preparation of ASA.
To determine if ethanoic anhydride is in excess for the preparation of ASA (acetylsalicylic acid), we need to compare the amount of ethanoic anhydride used in the laboratory to the amount required for the reaction.
First, let's calculate the mass of ethanoic anhydride used by using its density and volume:
Density of ethanoic anhydride = 1.092 g/mL
Volume of ethanoic anhydride used = 2.5 mL
Mass of ethanoic anhydride used = Density x Volume
Mass of ethanoic anhydride used = 1.092 g/mL x 2.5 mL = 2.73 g
Now, We need to compare this mass to the stoichiometric ratio of ethanoic anhydride to ASA in the reaction equation.
The balanced equation for the synthesis of ASA is:
2 C4H6O3 + C9H8O4 → 2 C6H8O2 + C2H4O2
From this equation, we can see that 2 moles of ethanoic anhydride (C4H6O3) react with 1 mole of ASA (C9H8O4).
Therefore, the molar ratio of ethanoic anhydride to ASA is 2:1.
So, the amount of ethanoic anhydride needed is half the amount of ASA produced.
If the amount of ASA produced in the laboratory is less than half the mass of the ethanoic anhydride used, then the ethanoic anhydride is in excess.
Since we don't know the amount of ASA produced, we can't directly determine if ethanoic anhydride is in excess. However, if the mass of ethanoic anhydride used (2.73 g) is greater than twice the mass of the ASA produced, then ethanoic anhydride is indeed in excess.
You need to determine the mass of ASA produced in the reaction to determine if ethanoic anhydride is in excess or not.