Question

Сh3-COONa→ch3cooh→ch3coo-c2h5oh

Answer 1

1)2CH3COONa+ H2SO4=Na2SO4 + 2CH3COOH 2) CH3COOH + C2H5OH= CH3COO-C2H5

Answer 2

The reaction you provided appears to be the saponification of an ester. Here's a step-by-step breakdown of the process:

1. CH3-COONa: This represents sodium acetate, which is the sodium salt of acetic acid (CH3COOH). It is commonly used as a reagent in organic chemistry.

2. CH3COOH: This represents acetic acid, which is a weak organic acid. In the presence of water, it can donate a proton (H+) to form the acetate ion (CH3COO-) and hydronium ion (H3O+).

3. CH3COO- + C2H5OH: In this step, the acetate ion (CH3COO-) reacts with ethanol (C2H5OH) to form ethyl acetate. This reaction is known as an esterification reaction and is catalyzed by an acid, such as sulfuric acid or hydrochloric acid. The reaction can be represented as follows:

   CH3COO- + C2H5OH → CH3COOC2H5 + H2O

   The product, ethyl acetate (CH3COOC2H5), is an ester with a fruity odor and is commonly used as a solvent.

So, the overall reaction you provided can be summarized as follows:

CH3-COONa + H2O → CH3COOH + NaOH CH3COOH + C2H5OH → CH3COOC2H5 + H2O

It's worth noting that the reaction may require specific reaction conditions, such as the presence of a catalyst or specific temperature, for the esterification step to occur effectively.