What is the balanced net ionic equation for the reaction of Al(OH)3(s) with HNO3(aq)? -...

Step 1
The Reaction between ${\displaystyle Al{\left(OH\right)}_3\text{and}HN{O}_3}$
${\displaystyle Al{\left(OH\right)}_3\left(s\right)+HN{O}_3\left(aq\right)\Rightarrow }$
Molecular Equation: In the molecular equation, reactant and product are represented as molecular substances even though they exist as ions in the solutions.
${\displaystyle Al{\left(OH\right)}_3\left(s\right)+3HN{O}_3\left(aq\right)\Rightarrow Al{\left(N{O}_3\right)}_3\left(aq\right)+3H_{2}O\left(l\right)}$
Step 2
Ionic Equation: Soluble ionic compounds exist as separate ions in solution.
${\displaystyle Al{\left(OH\right)}_3\left(s\right)+3H^{+}\left(aq\right)+3N{O}_3-\left(aq\right)\Rightarrow A{l}^{3+}\left(aq\right)+3N{O}_3-\left(aq\right)+3H_{2}O\left(l\right)}$
Net Ionic Equation: The ions that are common to both sides are canceled. These are called spectator ions and they do not participate in the chemical reaction.
As ${\displaystyle N{O}_3^{-}\left(aq\right)}$ is common to both sides, so it will get canceled.
${\displaystyle Al{\left(OH\right)}_3\left(s\right)+3H^{+}\left(aq\right)\Rightarrow A{l}^{3+}\left(aq\right)3H_{2}O\left(l\right)}$
${\displaystyle Al{\left(OH\right)}_3\left(s\right)+3H^{+}\left(aq\right)\Rightarrow A{l}^{3+}\left(aq\right)3H_{2}O\left(l\right)}$ is the net ionic equation
Step 3
Option (a) is the correct answer

The molecular reaction would be:
${\displaystyle Al\left(OH\right)3+HNO3\Rightarrow Al\left(NO3\right)3+3H2O}$
The net ionic equation would be as follows:
${\displaystyle Al\left(OH\right)3\left(s\right)+3H+\left(aq\right)\Rightarrow Al3+\left(aq\right)+3H2O\left(l\right)}$
Hope this helps.

Answer: $2Al(OH{)}_3(s)+6H^{+}(aq)+6N{O}_3^{⁻}(aq)\Rightarrow 2A{l}^{3+}+6N{O}_3^{⁻}+6H_{2}O(l)$ Explanation: 1) Start understanding the nature of the reactants, the kind of reaction that occurs, and the nature of the products: i) word equation: $\text{aluminium hydroxide}+\text{nitric acid}\Rightarrow \text{aluminum nitrate}+\text{water}$ ii) molecular equation, including the phases: $2Al(OH{)}_3(s)+6HN{O}_3(aq)\Rightarrow 2Al(N{O}_3{)}_3(aq)+6H_{2}O(l)$ iii) It is a neutralization reaction between an amphoteric compound, which is acting as a base, soluble in acid, and an acid in a aqueous solution, leading to the formation of a soluble salt and water in a double replacement reaction. 2) Separate the ionic aqueous compounds into its ions i) Reactant side: $6HN{O}_3(aq)=6H^{+}(aq)+6N{O}_3^{⁻}(aq)$ $Al(OH{)}_3$ (s) remains unchanged because it is in solid form ii) Product side $2Al(N{O}_3{)}_3(aq)=2A{l}^{3+}+6N{O}_3^{-}$ $H_{2}O(l)$ remains unchanged because it is in liquid form. 3) Write both sides, i.e. the complete reaction in the ionic form: $2Al(OH{)}_3(s)+6H^{+}(aq)+6N{O}_3^{-}(aq)\Rightarrow 2A{l}^{3+}+6N{O}_3^{-}+6H_{2}O(l)$ 4) Spectator ions Observe that there are not spectator ions. All the ions are participating. Hence, the net ionic equation is the same complete ionic equation.