Complete this equation for the reaction of NaHCO3(s) + HC2H3O2(aq) ---> Include phase symbols.
you get a salt, sodium acetate, and water, and carbon dioxide
Remember: acids+carbonates> salt + water+ CO2
(NH4)2CO3 + Cu(NO3)2 -> CuCO3 + 2NH4NO3
Getting rid of the spectator NH4 and NO3 ions, we are left with
Cu2+(aq) + CO32-(aq) -> CuCO3(s)
(NH2)2CO3 + Cu(NO3)2 -> CuCO3 + 2NH4NO3
Getting rid of the spectator NH4 and NO3 ions, we are left with
Cu2+(aq) + CO32-(aq) -> CuCO3(s)
(NH2)2CO3 + Cu(NO3)2 -> CuCO3 + 2NH4NO3
Getting rid of the spectator NH4 and NO3 ions, we are left with
Cu2+(aq) + CO32-(aq) -> CuCO3(s)
see
http://www.jiskha.com/display.cgi?id=1342376355
To complete the equation for the reaction of NaHCO3(s) + HC2H3O2(aq), we need to understand the nature of the reaction and properties of the reactants.
The reactants given are sodium bicarbonate (NaHCO3) in the solid state (s) and acetic acid (HC2H3O2) in the aqueous state (aq). Sodium bicarbonate is commonly known as baking soda, and acetic acid is the primary component of vinegar.
The reaction between sodium bicarbonate and acetic acid is an acid-base reaction, specifically a neutralization reaction. In this type of reaction, an acid reacts with a base to form a salt and water.
To write the balanced equation with phase symbols, we first need to know the chemical formula of the salt formed. This can be determined by identifying the ions present in sodium bicarbonate and acetic acid.
The ions in sodium bicarbonate are Na+ (sodium cation) and HCO3- (bicarbonate anion). The cation and anion combine to form NaHCO3.
The ions in acetic acid are H+ (hydrogen cation) and C2H3O2- (acetate anion). These ions combine to form HC2H3O2.
Now, we have enough information to write the balanced equation:
NaHCO3(s) + HC2H3O2(aq) → NaC2H3O2(aq) + H2O(l)
In this equation, (s) refers to solid, (aq) denotes aqueous (dissolved in water), and (l) indicates liquid.
Therefore, the balanced equation for the reaction of NaHCO3(s) + HC2H3O2(aq) is NaHCO3(s) + HC2H3O2(aq) → NaC2H3O2(aq) + H2O(l).