crealolobk

2022-01-14

What is the balanced net ionic equation for the reaction of ?
- $OAI{\left(OH\right)}_{3}\left(s\right)+3{H}^{+}\left(aq\right)⇒A{l}^{3+}\left(aq\right)+3{H}_{2}O\left(\mid \right)$
- $Al{\left(OH\right)}_{3}\left(s\right)+3{H}^{+}\left(aq\right)+3N{O}_{3}^{-}\left(aq\right)⇒A{l}^{3+}\left(aq\right)+3N{O}_{3}^{-}\left(aq\right)+3{H}_{2}O\left(\mid \right)$
- $Al{\left(OH\right)}_{3}\left(s\right)+{H}^{+}\left(aq\right)+N{O}_{3}^{-}\left(aq\right)⇒A{l}^{3+}\left(aq\right)+3{H}_{2}O\left(\mid \right)$
- $Al{\left(OH\right)}_{3}\left(s\right)+HN{O}_{3}\left(aq\right)⇒Al{\left(N{O}_{3}\right)}_{3}\left(aq\right)+3{H}_{2}O\left(\mid \right)$
- $Al{\left(OH\right)}_{3}\left(s\right)+{H}^{+}\left(aq\right)⇒A{l}^{3+}\left(aq\right)+{H}_{2}O\left(\mid \right)$

Jenny Sheppard

Expert

Step 1
The Reaction between
$Al{\left(OH\right)}_{3}\left(s\right)+HN{O}_{3}\left(aq\right)⇒$
Molecular Equation: In the molecular equation, reactant and product are represented as molecular substances even though they exist as ions in the solutions.
$Al{\left(OH\right)}_{3}\left(s\right)+3HN{O}_{3}\left(aq\right)⇒Al{\left(N{O}_{3}\right)}_{3}\left(aq\right)+3{H}_{2}O\left(l\right)$
Step 2
Ionic Equation: Soluble ionic compounds exist as separate ions in solution.
$Al{\left(OH\right)}_{3}\left(s\right)+3{H}^{+}\left(aq\right)+3N{O}_{3}-\left(aq\right)⇒A{l}^{3+}\left(aq\right)+3N{O}_{3}-\left(aq\right)+3{H}_{2}O\left(l\right)$
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 $N{O}_{3}^{-}\left(aq\right)$ is common to both sides, so it will get canceled.
$Al{\left(OH\right)}_{3}\left(s\right)+3{H}^{+}\left(aq\right)⇒A{l}^{3+}\left(aq\right)3{H}_{2}O\left(l\right)$
$Al{\left(OH\right)}_{3}\left(s\right)+3{H}^{+}\left(aq\right)⇒A{l}^{3+}\left(aq\right)3{H}_{2}O\left(l\right)$ is the net ionic equation
Step 3
Option (a) is the correct answer

Durst37

Expert

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

alenahelenash

Expert

Answer: $2Al\left(OH{\right)}_{3}\left(s\right)+6{H}^{+}\left(aq\right)+6N{O}_{3}^{⁻}\left(aq\right)⇒2A{l}^{3+}+6N{O}_{3}^{⁻}+6{H}_{2}O\left(l\right)$ Explanation: 1) Start understanding the nature of the reactants, the kind of reaction that occurs, and the nature of the products: i) word equation: $\text{aluminium hydroxide}+\text{nitric acid}⇒\text{aluminum nitrate}+\text{water}$ ii) molecular equation, including the phases: $2Al\left(OH{\right)}_{3}\left(s\right)+6HN{O}_{3}\left(aq\right)⇒2Al\left(N{O}_{3}{\right)}_{3}\left(aq\right)+6{H}_{2}O\left(l\right)$ 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: $6HN{O}_{3}\left(aq\right)=6{H}^{+}\left(aq\right)+6N{O}_{3}^{⁻}\left(aq\right)$ $Al\left(OH{\right)}_{3}$ (s) remains unchanged because it is in solid form ii) Product side $2Al\left(N{O}_{3}{\right)}_{3}\left(aq\right)=2A{l}^{3+}+6N{O}_{3}^{-}$ ${H}_{2}O\left(l\right)$ remains unchanged because it is in liquid form. 3) Write both sides, i.e. the complete reaction in the ionic form: $2Al\left(OH{\right)}_{3}\left(s\right)+6{H}^{+}\left(aq\right)+6N{O}_{3}^{-}\left(aq\right)⇒2A{l}^{3+}+6N{O}_{3}^{-}+6{H}_{2}O\left(l\right)$ 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.