NaOH+H3COOC2H5 --> CH3COONA+C2H5OH
To make the components more clear.
(Na-OH)+(H3COO-C2H5) --> (CH3COO-NA)+(C2H5-OH)
When sodium acetate reacts with sodium hydroxide, a double displacement reaction occurs. The products of the reaction are sodium hydroxide and sodium acetate. The balanced chemical equation for this reaction is: CH3COONa + NaOH → CH3COONa + NaOH
When Sodium Acetate is mixed with NaOH, a double displacement reaction occurs, leading to the formation of water and sodium hydroxide, along with sodium acetate. The reaction can be represented as follows: CH3COONa + NaOH → CH3COONa + H2O The sodium acetate remains in the solution, while water and sodium hydroxide are formed as byproducts.
Sodium Acetate Can be fond in 2 forms. Either anhydrous or trihydrate. Oxidation reaction with anhydrous form is easier than trihydrate form. First form has reaction similar to that of Oxidation of Acetic Acid. Trihydrate form is a bit more complex and I'm still loking into it
The reaction between C8H5O4K and NaOH will produce potassium salicylate (C7H5KO3) and water. The balanced equation is: C8H5O4K + NaOH → C7H5KO3 + H2O.
The reaction between ethyl iodide and alcoholic potash (potassium hydroxide dissolved in alcohol) results in the formation of ethyl alcohol, potassium iodide, and potassium ethoxide. The chemical equation for this reaction can be written as: C2H5I + KOH → C2H5OH + KI + KOC2H5
When sodium acetate reacts with sodium hydroxide, a double displacement reaction occurs. The products of the reaction are sodium hydroxide and sodium acetate. The balanced chemical equation for this reaction is: CH3COONa + NaOH → CH3COONa + NaOH
The reaction between ethyl 4-aminobenzoate and NaOH would involve the amine group of the 4-aminobenzoate being deprotonated by the strong base NaOH. This would result in the formation of the conjugate base of the amine group and water as a byproduct. The equation for this reaction can be represented as follows: Ethyl 4-aminobenzoate + NaOH → Ethyl 4-aminobenzoate-Na+ + H2O
When Sodium Acetate is mixed with NaOH, a double displacement reaction occurs, leading to the formation of water and sodium hydroxide, along with sodium acetate. The reaction can be represented as follows: CH3COONa + NaOH → CH3COONa + H2O The sodium acetate remains in the solution, while water and sodium hydroxide are formed as byproducts.
The reaction between sodium hydroxide (NaOH) and acetic acid (CH3COOH) forms sodium acetate (CH3COONa) and water (H2O). The balanced chemical equation is: CH3COOH + NaOH -> CH3COONa + H2O.
The balanced reaction for the reaction between HCl and NaOH is: HCl + NaOH -> NaCl + H2O This equation is already balanced as it shows the conservation of mass and charge.
The balanced equation for the reaction between acetic acid (HC2H3O2) and sodium hydroxide (NaOH) is: HC2H3O2 + NaOH → NaC2H3O2 + H2O
The balanced equation for the reaction between amidosulfuric acid (NH2SO3H) and sodium hydroxide (NaOH) is: NH2SO3H + NaOH → NaHSO3 + H2O
The reaction of acetic acid and sodium hydroxide will form sodium acetate and water. The chloroform is not involved in the reaction and will remain unchanged. The balanced chemical equation for the reaction is: CH3COOH (acetic acid) + NaOH (sodium hydroxide) -> CH3COONa (sodium acetate) + H2O (water)
The reaction between aqueous acetic acid (CH3COOH) and aqueous sodium hydroxide (NaOH) forms water (H2O) and sodium acetate (CH3COONa). The balanced chemical equation is: CH3COOH + NaOH -> H2O + CH3COONa
the equation for sodium acetate with water is NaC2H3O2+2(H2O)=Na+C2H3O2(solid).
To make Sodium Acetate, you would typically mix one mole of acetic acid (CH3COOH) with one mole of sodium hydroxide (NaOH). This will result in the formation of one mole of sodium acetate (CH3COONa) along with water. The balanced chemical equation for this reaction is CH3COOH + NaOH -> CH3COONa + H2O.
The reaction of ethanol with NaOH and iodine will yield iodoethane (ethyl iodide) as the product. The alcohol group in ethanol will be replaced by the iodine atom in the presence of NaOH.