Question

solid aluminum hydroxide reacts with a solution of hydrobromic acid. write a balanced molecular equation and a balanced net ionic equation, including the states of each species.
1st attempt
part 1 (1 point)
balanced molecular equation:
part 2 (1 point)
net ionic equation:

Explanation:

Step1: Write the reactants and products

Aluminum hydroxide ($Al(OH)_3(s)$) reacts with hydrobromic acid ($HBr(aq)$) to form aluminum bromide ($AlBr_3(aq)$) and water ($H_2O(l)$). The un - balanced molecular equation is $Al(OH)_3(s)+HBr(aq)
ightarrow AlBr_3(aq) + H_2O(l)$.

Step2: Balance the molecular equation

We need to balance the number of atoms on both sides. There are 3 bromine atoms in $AlBr_3$, so we need 3 moles of $HBr$. Also, to balance the hydrogen and oxygen atoms, we need 3 moles of $H_2O$. The balanced molecular equation is $Al(OH)_3(s)+3HBr(aq)
ightarrow AlBr_3(aq)+3H_2O(l)$.

Step3: Write the total ionic equation

$Al(OH)_3(s)$ is a solid and does not dissociate. $HBr(aq)$ is a strong acid and dissociates completely into $H^+(aq)$ and $Br^-(aq)$. $AlBr_3(aq)$ dissociates into $Al^{3 +}(aq)$ and $3Br^-(aq)$, and $H_2O(l)$ is a non - electrolyte. The total ionic equation is $Al(OH)_3(s)+3H^+(aq)+3Br^-(aq)
ightarrow Al^{3+}(aq)+3Br^-(aq)+3H_2O(l)$.

Step4: Write the net ionic equation

Cancel out the spectator ions ($Br^-$ in this case). The net ionic equation is $Al(OH)_3(s)+3H^+(aq)
ightarrow Al^{3+}(aq)+3H_2O(l)$.

Answer:

Part 1: $Al(OH)_3(s)+3HBr(aq)
ightarrow AlBr_3(aq)+3H_2O(l)$
Part 2: $Al(OH)_3(s)+3H^+(aq)
ightarrow Al^{3+}(aq)+3H_2O(l)$