What is the minimum of acetic anhydride (102.1 g/mol) required to react completely with the 2.94 grams of salicylic acid? What v

Question

2 years ago

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olume of acetic anhydride is this? (Density = 1.082 g/mL) Show your work. Include UNITS and round your final answer appropriately

1 answer:

Lady bird [3.3K] · Answer 1 · 2022-09-17T08:57:41+03:00

Answer:

a) the minimun of acetic anhydride required for the reaction is 2.175 g (CH3CO)2O

b) V acetic anhydride = 2.010 mL

Explanation:

C6H4OHCOOH + (CH3CO)2O ↔ C9H8O4 + C2H4O2

⇒ mol salicylic acid = 2.94 g C6H4OHCOOH * ( mol C6H4OHCOOH / 138.121 g ) = 0.0213 mol C6H4OHCOOH

⇒ mol acetic anhydride = 0.0213 mol C6H4OHCOOH * ( mol (CH3CO)2O / mol C6H4OHCOOH ) = 0.0213 mol (CHECO)2O

⇒ g acetic anhydride = 0.0213 mol * ( 102.1 g/mol ) = 2.175 g CH3CO)2O

b) V = 2.175 g (CH3CO)2 * ( mL / 1.082 g ) = 2.010 mL (CH3CO)2O