You're expressing those two reactants as ions. The chemical equation for the reaction between barium and nitrogen is: 3Ba + N2 --> Ba3N2. Or, if you like, the net ionic equation for the formation of the insoluble precipitate barium nitride is: 3Ba+2 + 2N-3 --> Ba3N2.
2(NH4)3PO4 + 3Ba(OH)2 --> 6NH4OH + Ba3(PO4)2
3Ba(BrO3)2 + 2Na3PO4 -> Ba3(PO4)2 + 6NaBrO3
The molecular equation is 3Ba(NO3)2(aq) + 2(NH4)3PO4(aq) ==> Ba3(PO4)2(s) + 6NH4NO3(aq)The spectator ions are NH4^+ and NO3^-
The solubility of barium is somewhat mixed---it can be soluble with notoriously insoluble things like hydroxide, and it can be insoluble with other somewhat insoluble anions like sulfate. However, chromate is one anion that is almost always insoluble unless it is paired with an alkali metal. So mixing these two compounds will give you a BaCrO4 precipitate.
Assuming cobalt(II) nitrate: Co(NO3)2 + Na2CO3 --> CoCO3 + 2NaNO3
2Na3PO4 + 3Ba(NO3)2 → Ba3(PO4)2 + 6NaNO3
You're expressing those two reactants as ions. The chemical equation for the reaction between barium and nitrogen is: 3Ba + N2 --> Ba3N2. Or, if you like, the net ionic equation for the formation of the insoluble precipitate barium nitride is: 3Ba+2 + 2N-3 --> Ba3N2.
2H3PO4(aq) + 3Ba(OH)2(aq) > Ba3(PO4)2(aq) + 6H2O(l)
2(NH4)3PO4 + 3Ba(OH)2 --> 6NH4OH + Ba3(PO4)2
3Ba(NO3)2(aq) + 2(NH4)3PO4(aq) --> Ba3(PO4)2(s) + 6NH4NO3(aq)
2Na3(po4)(aq) +3Ba(C2H3O2)2(AQ)=Ba3(Po4)2(s) + 6NaO2C2H3(aq)
3Ba(BrO3)2 + 2Na3PO4 -> Ba3(PO4)2 + 6NaBrO3
(NH4)3 2PO4 + 3Ba (OH)2 = 3NH4 2OH + Ba3 (PO4)2
2H3C6H5O7 + 3Ba(OH)2 -> 6H2O + Ba3(C6H5O7)2
Barium reacts with nitrogen to produce Ba3N2. 3Ba + 2N -> Ba3N2 The is a synthesis reaction.
2K 3 PO 4 + 3Ba(NO 3 ) 2 -----> 6KNO 3 + Ba 3 (PO 4 ) 2