The net ionic equation for the given reaction is H+ (aq) + OH- (aq) β H2O (l)
The net ionic equation for Na^+ + Cl^- is Na^+ + Cl^- β NaCl, which represents the formation of sodium chloride when Na^+ and Cl^- ions combine. This equation shows the reactants and products without including spectator ions that do not participate in the reaction.
The reaction is already balanced as it is written: Clβ + 2Kl β 2KCl + Clβ. Each side of the reaction has the same number of atoms for each element.
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) -> H2O(l) + Na+(aq) + Cl-(aq)
The formula for tin ions depends on their charge. Tin can exist as Sn2+ ions (called stannous ions) or Sn4+ ions (called stannic ions). The formula for stannous ions is Sn2+ and for stannic ions is Sn4+.
The symbol for the tin IV ion is Sn4+.
Na+ and Cl- are spectator ions.
Sn4+ is fully oxidised, Sn2+ only half
The positive ion for tin IV is Sn^4+, and the chloride ion is Cl^-. Therefore, when tin IV chloride is formed, it would be written as SnCl4.
FeCl3
Here is a covalent bond.
Hydrochloric acid and sodium hydroxide yield salt and water H+ + Cl- + Na+ + OH- --> Na+ + Cl- + H2OComment:In solutions you better leave unchanged ions ( Cl- and Na+) out of the balanced equation: called to be 'tribune ions' (people on the tribune don't take part in the 'match'):H+ + OH- --> H2O This looks simpler than: H+ + Cl - + Na + + OH- --> Na + + Cl - + H2O
Sn4+ is the symbol for Tin(IV), that is, the element tin with a oxidation state of 4.
Sn4 0dz
The net ionic equation for the given reaction is H+ (aq) + OH- (aq) β H2O (l)
2 cl are needed
CaCl2