# Question #a9d9d

Dec 22, 2017

Molecular $N {a}_{2} C {O}_{3} + 2 H C l \to 2 N a C l + {H}_{2} O + C {O}_{2}$

Complete Ionic $2 N {a}^{+} \left(a q\right) + C {O}_{3}^{2 -} \left(a q\right) + 2 {H}^{+} \left(a q\right) + 2 C {l}^{-} \left(a q\right) \to 2 N {a}^{+} \left(a q\right) + 2 C {l}^{-} \left(a q\right) + {H}_{2} O \left(l\right) + C {O}_{2} \left(g\right)$

Net Ionic $2 N {a}^{+} \left(a q\right) + 2 {H}^{+} \left(a q\right) \to 2 N a C l \left(a q\right)$

#### Explanation:

The molecular equation is the overall balanced equation, including states:

$N {a}_{2} C {O}_{3} \left(a q\right) + 2 H C l \left(a q\right) \to 2 N a C l \left(a q\right) + {H}_{2} O \left(l\right) + C {O}_{2} \left(g\right)$

To write the complete ionic equation , you break all (aq) compounds from the molecular equation into their individual ions:

$2 N {a}^{+} \left(a q\right) + C {O}_{3}^{2 -} \left(a q\right) + 2 {H}^{+} \left(a q\right) + 2 C {l}^{-} \left(a q\right) \to 2 N {a}^{+} \left(a q\right) + 2 C {l}^{-} \left(a q\right) + {H}_{2} O \left(l\right) + C {O}_{2} \left(g\right)$

We know that $C {O}_{3}^{2 -}$ has a 2- charge, because it is bonded to two $N {a}^{+}$ ions, which each have a positive charge. Thus it must have a 2- charge to balance out the 2+ and form the neutral $N {a}_{2} C {O}_{3}$

To write the net ionic equation , you remove all of the spectator ions (ions which do not take part in the reaction) from your complete ionic equation:

$2 N {a}^{+} \left(a q\right) + 2 {H}^{+} \left(a q\right) \to 2 N a C l \left(a q\right)$