Question

Question #a9d9d

Answer 1

Molecular `Na_2CO_3 + 2HCl -> 2NaCl + H_2O + CO_2`

Complete Ionic `2Na^+(aq) + CO_3^(2-)(aq) + 2H^+(aq) + 2Cl^(-)(aq) -> 2Na^+(aq) + 2Cl^(-)(aq) + H_2O (l) + CO_2 (g)`

Net Ionic `2Na^+(aq) + 2H^+(aq) -> 2NaCl(aq)`

Explanation:

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

`Na_2CO_3(aq) + 2HCl(aq) -> 2NaCl(aq) + H_2O(l) + CO_2(g)`

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

`2Na^+(aq) + CO_3^(2-)(aq) + 2H^+(aq) + 2Cl^(-)(aq) -> 2Na^+(aq) + 2Cl^(-)(aq) + H_2O (l) + CO_2 (g)`

We know that `CO_3^(2-)` has a 2- charge, because it is bonded to two `Na^+` ions, which each have a positive charge. Thus it must have a 2- charge to balance out the 2+ and form the neutral `Na_2CO_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:

`2Na^+(aq) + 2H^+(aq) -> 2NaCl(aq)`