HC2H3O2(aq) + KOH(aq) -> H2O(l) + KC2H3O2(aq)
It's already balanced, since there are two acetates, two Hydrogens on each side, and 1 oxygen on both sides :)
The equation is already balanced! :)
Like so:
HCl + KOH --> KCl + H2O
It's a neutralization reaction.
H2C2O4 + 2KOH -> 2H2O + K2C2O4
The equation is KOH + HC2H3O2 --> KC2H3O2 + H2O
CaO + H20 ---> Ca(OH)2
HCl + KOH -----> KCl + H2O
KCl
This equation is Al2O3 + 6 HCl = 2 AlCl3 + 3 H2O.
Cs2CO3 + 2HCl--->2CsCl+CO2+H2O
HCl + NaOH = H2O + NaCl is already balanced.
The balanced equation is 2HCl + K2O -> H2O + 2KCl.
3
This equation is Al2O3 + 6 HCl = 2 AlCl3 + 3 H2O.
BaOH + HCl -> BaCl + H2O
Cs2CO3 + 2HCl--->2CsCl+CO2+H2O
HCl + NaOH = H2O + NaCl is already balanced.
The balanced equation is 2HCl + K2O -> H2O + 2KCl.
3
This equation is HCl + NaOH -> NaCl + H2O.
The balanced equation for the reaction between aqueous sodium hypochlorite (NaOCl) and aqueous hydrochloric acid (HCl) is: NaOCl + HCl → NaCl + Cl2 + H2O
The chemical equation is:PbO2 + 4 HCl = PbCl2 + Cl2 + 2 H2O
HCl + NaOH --> NaCl + H2O is balanced as you wrote it.But, since sodium is always soluble in water at temperatures below the boiling point of the solution, the net ionic equation for the reaction at temperatures lower than the boiling point would actually be: HCl + OH- --> Cl- + H2OOr, if the hydrochloric acid was already in solution, then simplyH+ + OH- ---> H2O
HCl + NaOH --> NaCl + H2O
Pcl3 + 3h2o-------> h3po3 + 3 hcl