Problem 43 - Chemistry The Molecular Nature of Matter

Question

Write equations for the reaction of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) with (a) a strong acid, and (b) a strong base.

Step-by-Step Solution

Answer

a) \(\mathrm{Al}_2\mathrm{O}_3 + 6\mathrm{HCl} \rightarrow 2\mathrm{AlCl}_3 + 3\mathrm{H}_2\mathrm{O}\) b) \(\mathrm{Al}_2\mathrm{O}_3 + 2\mathrm{NaOH} + 3\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{NaAlO}_2 + 4\mathrm{H}_2\mathrm{O}\)

Step 1: Reaction with a Strong Acid

The reaction between aluminum oxide \(\mathrm{Al}_2\mathrm{O}_3\) and a strong acid, such as hydrochloric acid \(\mathrm{HCl}\), results in the formation of aluminum chloride \(\mathrm{AlCl}_3\) and water. We write the balanced chemical equation: \[\mathrm{Al}_2\mathrm{O}_3(s) + 6\mathrm{HCl}(aq) \rightarrow 2\mathrm{AlCl}_3(aq) + 3\mathrm{H}_2\mathrm{O}(l)\]

Step 2: Reaction with a Strong Base

Aluminum oxide reacts with a strong base like sodium hydroxide \(\mathrm{NaOH}\) to form sodium aluminate \(\mathrm{NaAlO}_2\) and water. The balanced chemical equation for this reaction is: \[\mathrm{Al}_2\mathrm{O}_3(s) + 2\mathrm{NaOH}(aq) + 3\mathrm{H}_2\mathrm{O}(l) \rightarrow 2\mathrm{NaAlO}_2(aq) + 4\mathrm{H}_2\mathrm{O}(l)\]

Key Concepts

Understanding Chemical Equations

Chemical equations are vital for describing the substances involved in chemical reactions, their quantities, and the products formed. The equation represents with symbols and formulas, the identities and relative amounts of the reactants and products. For example, in the reaction of aluminum oxide with a strong acid or a base, it's essential to balance the equation to respect the law of conservation of mass, meaning the number of atoms of each element must be the same on both sides of the equation.

When looking at the equation \[\mathrm{Al}_2\mathrm{O}_3(s) + 6\mathrm{HCl}(aq) \rightarrow 2\mathrm{AlCl}_3(aq) + 3\mathrm{H}_2\mathrm{O}(l)\], the coefficients (like the '6' in front of HCl) ensure that there are equal numbers of each type of atom on the reactants side (left) as on the products side (right). Balancing chemical equations requires practice and understanding of stoichiometry, which is the quantitative study of reactants and products in a chemical reaction.

Reaction with Strong Acid

Aluminum oxide reacts with strong acids to form a salt and water, a process known as neutralization. The reaction of aluminum oxide \(\mathrm{Al}_2\mathrm{O}_3\) with hydrochloric acid \(\mathrm{HCl}\) is a prime example. In a neutralization reaction, the acid donates protons (H+) and the base, in this case, aluminum oxide, accepts them, which results in the formation of water \(\mathrm{H}_2\mathrm{O}\) and a salt, aluminum chloride \(\mathrm{AlCl}_3\).

It's intriguing to note that aluminum oxide can act as an acid or base, showcasing its amphoteric nature. The product, aluminum chloride, is often used in various applications in industries and laboratories, proving the practical significance of understanding such reactions for students planning to venture into scientific fields.

Reaction with Strong Base

In contrast to its reaction with acids, aluminum oxide \(\mathrm{Al}_2\mathrm{O}_3\) when mixed with a strong base such as sodium hydroxide \(\mathrm{NaOH}\), undergoes a different type of chemical reaction. The aluminum oxide acts as an acid and reacts with the base to form sodium aluminate \(\mathrm{NaAlO}_2\) and water. This process is an example of an acid-base reaction where the aluminum oxide exhibits its amphoteric characteristics, meaning it can react with both acids and bases.

The reaction \[\mathrm{Al}_2\mathrm{O}_3(s) + 2\mathrm{NaOH}(aq) + 3\mathrm{H}_2\mathrm{O}(l) \rightarrow 2\mathrm{NaAlO}_2(aq) + 4\mathrm{H}_2\mathrm{O}(l)\]is significant because sodium aluminate is used in water treatment and paper manufacturing. Understanding these reactions allows students to predict the outcomes of experiments and grasp the usage of aluminum oxide in industrial processes.