ok so you have 1.76g of salicyic acid(138.12 g/mol) and with 1.11 g of acetic anhydride(102.10 g/mol) a student obtained 1.12g of acetylsalicylic acid( 180.17 g/mol) what is the percentage yield?

Do you know how to determine which is the limiting reagent? Do that.

Convert grams of the limiting reagent (LR)to mols of the LR then use the coefficients in the balanced equation to convert mols LR to mols of the product. That is the theoretical yield (TY).

Then %yield = (actual yield/TY)*100 = ?

To calculate the percentage yield, we need to compare the actual yield (1.12 g) to the theoretical yield. The theoretical yield can be calculated using stoichiometry.

First, we need to determine the balanced chemical equation for the reaction between salicylic acid and acetic anhydride to form acetylsalicylic acid.

The reaction equation is as follows:
C7H6O3 (salicylic acid) + C4H6O3 (acetic anhydride) → C9H8O4 (acetylsalicylic acid) + CH3COOH (acetic acid)

From the equation, we can see that the molar ratio between salicylic acid and acetylsalicylic acid is 1:1. This means that for every 1 mole of salicylic acid, we should obtain 1 mole of acetylsalicylic acid.

Now, let's calculate the number of moles of salicylic acid and acetic anhydride we have:

Number of moles of salicylic acid = mass of salicylic acid / molar mass of salicylic acid
Number of moles of salicylic acid = 1.76 g / 138.12 g/mol = 0.0127 mol

Number of moles of acetic anhydride = mass of acetic anhydride / molar mass of acetic anhydride
Number of moles of acetic anhydride = 1.11 g / 102.10 g/mol = 0.0109 mol

Since the molar ratio between salicylic acid and acetylsalicylic acid is 1:1, the number of moles of acetylsalicylic acid formed should be equal to the number of moles of salicylic acid used.

Therefore, the theoretical yield of acetylsalicylic acid is 0.0127 mol.

Now we can calculate the theoretical yield in grams:

Theoretical yield = number of moles of acetylsalicylic acid × molar mass of acetylsalicylic acid
Theoretical yield = 0.0127 mol × 180.17 g/mol = 2.29 g

Finally, we can calculate the percentage yield:

Percentage yield = (actual yield / theoretical yield) × 100%
Percentage yield = (1.12 g / 2.29 g) × 100% ≈ 48.9 %

Therefore, the percentage yield of the reaction is approximately 48.9%.