CH3COOH + NH3 ==> CH3COO- + NH4+
CH3COOH + NH3 ==> CH3COO- + H+ + NH3 ==> CH3COO- + NH4+
Since CH3COOH is a weak acid and NH3 is a weak base, I would not treat neither CH3COO- not NH3 as a spectator.
2007-10-15 19:21:59
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answer #1
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answered by Hahaha 7
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i think of you're lacking something on your equation. Sodium metallic, Na(s), plus HCl will consequence in a redox reaction, no longer a metathesis reaction: 2Na(s) + 2HCl ----> 2 NaCl + H2(g) that's if the solvent is HCl. If the solvent is water, then you gets sodium hydroxide forming too. once you're reacting a sodium salt like NaOH with HCl, a extra attainable occasion to apply for ionic equations, the molecular equation would be: NaOH(aq) + HCl(aq) ----> H2O(l) + NaCl(aq) This element shows all the molecules as though they weren't ionizing in answer. it rather is actual ordinary that they do ionize nonetheless, so a extra precise thank you to depict the reaction is the ionic equation, which shows the species as they exist interior the reaction medium (water, for that reason). the great ionic equation shows all the ions of their solvated, dissociated state, and any insoluble and/or covalently bonded reactants or products that are shaped besides: Na+ + OH- + H+ + Cl- ----> H2O(l) + Na+ + Cl- the internet ionic equation leaves out something that occurs as its solvated self in the two the reactants and the products. for that reason, the Cl- and Na+ ions are interior a similar state on the two element of the reaction, so which you would be able to easily... cancel them out. the only "product" shaped is the recent, non-ionic molecule of water, and the only ions in touch in its formation are the H+ and the Cl- ions: OH- + H+ -----> H2O(l) you could write Na(s) + HCl in an ionic vogue, despite the fact that it might no longer likely describe the reaction o.k.. Ionic equations are inteded to describe metathesis reactions extra suitable than they are meant to describe redox reactions. nevertheless, the great and internet ionic equations for Na + HCl would be: 2Na(s) + 2H+(aq) + 2Cl-(aq) -----> H2(g) + 2Na+(aq) + 2Cl-(aq) 2Na(s) + 2H+(aq) -----> H2(g) + Na+
2016-12-29 10:39:25
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answer #2
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answered by guisinger 3
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