Give the total ionic and net ionic equation NH4NO3(aq) + NaOH(aq) ----> NH3(s) +

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when you use solid NaOH, only nitrate is a spectator (NH4+ & NO3-) (aq) + NaOH(s)--->NH3(g) +H2O (l)+ (Na+ & NO3-)(aq) gives a net of: (NH4+ (aq) + NaOH(s) ---> NH3(g) +H2O (l) + Na+ (aq) =================================== as an edit: if you add solid NaOH to the NH4NO3 solution , your net ionic includes NaOH (s) releasing its Na+ (aq) as a product: (NH4+ (aq) + NaOH(s) ---> NH3(g) +H2O (l) + Na+ (aq) only if you dissolve the NaOH(s) ahead of time, does Na+ come out of the net ionic equation of NaOH(aq) reacting with NH4NO3(aq): (NH4+ (aq) + (Na+ & OH-)(aq) ---> NH3(g) +H2O (l) + Na+ (aq) becomes: (NH4+ (aq) & OH-(aq) ---> NH3(g) +H2O (l)

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