NaC2H3O2 (aq) + HCl (aq) = NaCl (aq) + ?
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HC2H3O2.
Just a double decomposition (or double displacement).
To determine the complete and balanced equation for the reaction between NaC2H3O2 (sodium acetate) and HCl (hydrochloric acid), we must first understand the properties and behavior of these compounds.
NaC2H3O2 is a salt of sodium, derived from acetic acid. It dissociates in water to produce sodium ions (Na+) and acetate ions (C2H3O2-).
HCl is a strong acid that dissociates completely in water, yielding hydrogen ions (H+) and chloride ions (Cl-).
Now let's write the balanced equation step by step:
1. Write the chemical formula for sodium acetate and hydrochloric acid:
NaC2H3O2 (aq) + HCl (aq)
2. Identify the ions present in each compound when they dissociate in water:
NaC2H3O2 (aq): Na+ (aq) + C2H3O2- (aq)
HCl (aq): H+ (aq) + Cl- (aq)
3. Determine the balanced equation by swapping the ions between the compounds:
NaC2H3O2 (aq) + HCl (aq) → Na+ (aq) + Cl- (aq) + C2H3O2- (aq)
4. Finally, we combine the sodium and chloride ions to form NaCl (sodium chloride), which is soluble in water:
Na+ (aq) + Cl- (aq) + C2H3O2- (aq) → NaCl (aq) + C2H3O2- (aq)
Therefore, the balanced equation for the reaction between NaC2H3O2 (sodium acetate) and HCl (hydrochloric acid) is:
NaC2H3O2 (aq) + HCl (aq) → NaCl (aq) + C2H3O2- (aq)
In this reaction, sodium acetate reacts with hydrochloric acid to form sodium chloride and acetic acid.