Question

A metal \(X\) reacts with aqueous \(\mathrm{NaOH}\) solution to form \(Y\) and a highly inflammable gas. Solution \(Y\) is heated and \(\mathrm{CO}_{2}\) is poured through it. \(Z\) precipitates out and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is formed. \(Z\) on heating gives \(\mathrm{Al}_{2} \mathrm{O}_{3}\). Identify \(X, Y\) and \(Z\). $$\begin{array}{lll} {\boldsymbol{X}} & {\boldsymbol{Y}} & {\boldsymbol{Z}} \\ (a) \mathrm{Al} &\mathrm{NaAlO}_{2} & \mathrm{Al}(\mathrm{OH})_{3} \\ (b) \mathrm{Al}_{2} \mathrm{O}_{3} & \mathrm{NaAlO}_{2} & \mathrm{Al}_{2} \mathrm{CO}_{3} \\ (c) \mathrm{Al}_{2} \mathrm{O}_{3} & {\left[\mathrm{Na}_{2} \mathrm{AlO}_{2}\right]^{+} \mathrm{OH}^{-}} & \mathrm{Al}(\mathrm{OH})_{3} \\\ (d) \mathrm{Al} & \mathrm{Al}(\mathrm{OH})_{3} & \mathrm{Al}_{2} \mathrm{O}_{3} \end{array}$$

Short answer

Metal X is Aluminum (Al), Compound Y is sodium tetrahydroxoaluminate(III) (NaAl(OH)4 or NaAlO2), and Compound Z is Aluminum hydroxide (Al(OH)3).

Step-by-step solution

Determine Metal X

The metal X reacts with NaOH to form Y and a highly inflammable gas. The highly inflammable gas could be hydrogen, which is produced when aluminum (Al) reacts with NaOH. The reaction is: \(2Al(s) + 2NaOH(aq) + 6H_2O(l) = 2NaAl(OH)_4(aq) + 3H_2(g)\). The metal X is therefore aluminum (Al).

Identify Compound Y

Compound Y is formed alongside hydrogen gas when aluminum reacts with NaOH. The result of this reaction is sodium aluminate, which in aqueous solution is often written as sodium tetrahydroxoaluminate(III), NaAl(OH)_4. Compound Y is therefore NaAl(OH)_4, which may also be represented in some contexts as NaAlO2 (sodium aluminate) in its anhydrous form.

Find the Precipitate Z

When CO2 is passed through the solution of Y, it reacts to form a precipitate Z and Na2CO3. For aluminum salts, when CO2 is passed through their aqueous solutions, aluminum hydroxide precipitates out: \(2NaAl(OH)_4(aq) + CO2(g) = 2Al(OH)_3(s) + Na2CO3(aq)\). Therefore, precipitate Z is aluminum hydroxide, Al(OH)3.

Verify Z on heating gives Al2O3

When aluminum hydroxide is heated, it decomposes to form aluminum oxide and water: \(2Al(OH)_3(s) = Al2O3(s) + 3H2O(l)\). This step confirms that heating Z, which is Al(OH)3, yields Al2O3.

Key concepts

Aluminum Reaction with NaOH

The reaction between aluminum (Al) and sodium hydroxide (NaOH) is fascinating and critical in various industrial processes. When aluminum comes into contact with NaOH, especially in aqueous solution, it reacts vigorously to form sodium aluminate and hydrogen gas — a highly flammable byproduct. The balanced chemical equation for this reaction is:

\[2Al(s) + 2NaOH(aq) + 6H_2O(l) \rightarrow 2NaAl(OH)_4(aq) + 3H_2(g)\]

This equation shows solid aluminum (\(Al(s)\)) reacting with sodium hydroxide in water to form an aqueous sodium tetrahydroxoaluminate(III) (\(NaAl(OH)_4(aq)\)), releasing hydrogen gas (\(H_2(g)\)). This reaction underpins the idea that Al can react with bases as well as acids—a property classified under amphoteric behavior. Understanding this chemical behavior is crucial as it explains how aluminum can be used in various applications, from water treatment to the synthesis of other aluminum compounds.

Sodium Aluminate Formation

Sodium aluminate, commonly represented as \(NaAlO_2\), is the product of aluminum reacting with sodium hydroxide. In its hydrated form, it can be written as \(NaAl(OH)_4\). But for ease and considering its anhydrous form, the simplified notation is often preferred in chemistry discussions. The formation is of significant importance because it represents a step that transforms aluminum into a soluble form.

Sodium aluminate serves as an intermediate in various chemical processes, like water purification, where it helps in removing impurities by coagulation. Additionally, it operates as a source material for other aluminum compounds. The ability to form such compounds from aluminum is one of the metal's most valuable properties, especially in industrial chemistry, making this reaction not just a textbook example but a real-world application.

Aluminum Hydroxide Precipitation

Aluminum hydroxide precipitation is a critical phenomenon that stems from introducing \(CO_2\) to a solution of sodium aluminate. In the given context, after the formation of sodium aluminate (\(NaAl(OH)_4\)), passing \(CO_2\) through the solution leads to the following reaction:

\[2NaAl(OH)_4(aq) + CO_2(g) \rightarrow 2Al(OH)_3(s) + Na_2CO_3(aq)\]

As a result, aluminum hydroxide (\(Al(OH)_3\)), a gelatinous white precipitate, is formed. This kind of reaction is fundamental in explaining how a soluble aluminum base can be converted back into an insoluble form. It also illustrates a typical response of aluminum compounds with carbon dioxide. Precipitation reactions like these are not only critical for understanding chemical reactivity but also for practical applications like water treatment, where the removal of unwanted ions is necessary.

Chemical Decomposition of Al(OH)3

The chemical decomposition of aluminum hydroxide (\(Al(OH)_3\)) is a thermal process where heat causes the compound to break down into aluminum oxide (\(Al_2O_3\)) and water. Upon heating, the reaction is depicted as follows:

\[2Al(OH)_3(s) \rightarrow Al_2O_3(s) + 3H_2O(l)\]

This decomposition is essential as it shows the transformation of a hydroxide into an oxide, which in the case of aluminum, results in the production of aluminum oxide or alumina. Alumina is a significant material used in various industries, including the production of aluminum metal, abrasives, refractories, and as a filler for plastics. It is also important in the synthesis of other materials, such as synthetic sapphires for lasers and watches. Decomposition reactions like this are a crucial concept in chemistry, demonstrating the stability of compounds and the conditions under which they break down.