Formation of butyl propanoate and propyl methanoate

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The discussion focuses on the balanced equations for the formation of butyl propanoate and propyl methanoate, highlighting the reactants and the removal of water. For butyl propanoate, the reaction involves propanoic acid and butanol, producing butyl propanoate and water. In the case of propyl methanoate, the reactants are propanol and methanoic acid, yielding propyl methanoate and water. Participants confirm that the equations are balanced and express satisfaction with the clarity of the water removal process. Overall, the thread effectively illustrates the esterification reactions and their balanced equations.
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91. Write balanced equations to show the formation of each of the following compounds. Name the reactants in each case, and show clearly the removal of the water molecule.

a) butyl propanoate
b) propyl methanoate

Answers:

a) butyl propanoate- C7H14O2 = (CH3CH2COOCH2CH2CH2CH3)

C2H5COOH + C4H9OH ---> C7H14O2 + H2O

Propanoic acid + Butanol ---> butyl propanoate + water



b) propyl methanoate - C4H8O2 = (HCOOCH2CH2CH3)

C3H7OH + HCOOH ---> C4H8O2 + H2O

Propanol + methanoic acid ---> propyl methanoate + water



The equations are both balanced right now with the water molecules, but how do I show the removal of the water molecule and make them balanced again?
 
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You've already shown it.
 
Oh sweet!

Thanks.
 
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