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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.
Find a very form of your chemical reaction:
`Au + HBr -> AuBr_3 + H_2`
`2Au + 6 HBr -> 2AuBr_3 + 3 H_2`
You should have equal amounts of each element on either side of the reaction.
Split the ionic bonds:
`2Au + 6 H^+ + 6 Br^(-) -> 2Au^(3+) + 6 Br^(-) + 3H_2`
In a net ionic equation, you can cancel out any ions or molecules that are on both sides of the equation (spectator ions). In the reactants, there are 6 bromide ions. Likewise, in the products, there are 6 bromide ions. The bromide ions cancel each other out.
`2Au + 6H^+ -> 2Au^(3+) + 3H_2`
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