The chemical equation for the reaction of gold and hydrobromic acid can be expressed as:
`Au _(s) + HBr _(aq) -> AuBr_3 _(aq) + H_2 _(g) `
Balancing the reaction, we can have:
`2 Au _(s) + 6 HBr _(aq) -> 2 AuBr_3 _(aq) + 3 H_2 _(g) `
The next thing to do is to show the ionic forms of the species that are ions in the solution. This is also the complete ionic equation of the reaction of Au and HBr.
`2 Au _(s) + 6 H^(+) _(aq) + 6 Br^(-)_(aq)-> 2 Au ^(3+) + 6 Br^(-)_(aq)+ 3 H_2 _(g) `
To get the net ionic equation, you will remove (in other case, subtract) the spectator ions. These ions are those that are present in the reactants and product side. Since both sides of the reaction contain 6 moles of Br-, it can be removed.
`2 Au _(s) + 6 H^(+)_(aq)-> 2 Au ^(3+) _(aq) + 3 H_2 _(g) `
Take note that Au and H have different states in the reactant and product side.