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April 16, 2014

Homework Help: CHEM HELP

Posted by Yaw on Tuesday, November 27, 2012 at 1:34am.

I asked this already but for some reason the answer I got isnt being accepted online
Question:
In the laboratory, 3.50 grams of salicylic acid were reacted with 7.20 ml of acetic anhydride (density equal to 1.08 g/ml). Determine the limiting reagent.

I did:
salicylic acid + acetic anhydride → acetyl salicylic acid + acetic acid
C6H4(OH)COOH + (CH3CO)2O → CH3COOC6H4COOH + CH3COOH
moles C6H4(OH)COOH = (mass C6H4(OH)COOH) / (mole mass C6H4(OH)COOH)
moles C6H4(OH)COOH = (3.50 g) / (138.12354 g/mol)
moles C6H4(OH)COOH = 0.025396 mol
density (CH3CO)2O = [mass (CH3CO)2O] / [volume (CH3CO)2O]
or
mass (CH3CO)2O = [density (CH3CO)2O][volume (CH3CO)2O]
mass (CH3CO)2O = [1.08 g/mL][7.20 mL]
mass (CH3CO)2O = 7.776 g
moles (CH3CO)2O = [mass (CH3CO)2O] / [mole mass (CH3CO)2O]
moles (CH3CO)2O = [7.776 g] / [102.09024 g/mL]
moles (CH3CO)2O = 0.0761679 mol

So the answer would be salycic acid b/c its the lower number 0.025 mol

BUT my answer is incorrect apparently. Is here something I did wrong?

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