Write the net ionic equation for the precipitation of iron(II) sulfide from aqueous solution:
Fe^2+(aq) + S^2-(aq) ==> FeS(s)
2Fe3+(aq)+ 3S2-(aq)Fe2S3(s)
Fe^2+(aq) + H2S(aq) -> FeS(s) + 2H+(aq)
To write the net ionic equation for the precipitation of iron(II) sulfide from aqueous solution, we need to first identify the reactants and products involved in the reaction.
The reactants are iron(II) ions (Fe^2+) and sulfide ions (S^2-). In aqueous solution, these ions might be present as FeSO4 and Na2S, respectively.
The balanced chemical equation for the reaction can be written as follows:
FeSO4 + Na2S -> FeS + Na2SO4
Now, let's write the ionic equation by representing the reactants and products as their respective ions:
Fe^2+ (aq) + SO4^2- (aq) + 2Na^+ (aq) + S^2- (aq) -> FeS (s) + 2Na^+ (aq) + SO4^2- (aq)
The next step is to cancel out the spectator ions, which are the ions that appear on both sides of the equation and do not participate in the reaction. In this case, the Na^+ and SO4^2- ions are present on both sides and can be canceled out:
Fe^2+ (aq) + S^2- (aq) -> FeS (s)
Therefore, the net ionic equation for the precipitation of iron(II) sulfide is:
Fe^2+ (aq) + S^2- (aq) -> FeS (s)