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\(\mathbf{A}\) white solid \(X\) on heating gives a white solid \(Y\) and an acidic gas \(Z\). Gas \(Z\) is also given out when \(X\) reacts with an acid. The compound \(Y\) is also formed if canstic soda is left open in the atmosphere, \(X, Y\) and \(Z\) are\begin{aligned} &\mathbf{X}\\\ &\boldsymbol{Y} \quad \mathbf{Z}\\\ &\begin{array}{lll} \text { (a) } \mathrm{NaHCO}_{3} & \mathrm{Na}_{2} \mathrm{CO}_{3} & \mathrm{CO}_{2} \end{array}\\\ &\begin{array}{lll} \text { (b) } \mathrm{Na}_{2} \mathrm{CO}_{3} & \mathrm{NaOH} & \mathrm{CO}_{2} \end{array}\\\ &\text { (c) } \mathrm{Na}_{2} \mathrm{CO}_{3}\\\ &\mathrm{NaHCO}_{3} \quad \mathrm{CO}_{2}\\\ &\begin{array}{lll} \text { (d) } \mathrm{NaOH} & \mathrm{NaHCO}_{3} & \mathrm{CO}_{2} \end{array} \end{aligned}

Short Answer

Expert verified
Substance X is sodium bicarbonate (NaHCO3), Y is sodium carbonate (Na2CO3), and Z is carbon dioxide (CO2), so the correct option is (a).

Step by step solution

01

Identify the substances involved in the chemical process

To discern the identity of compounds X, Y, and Z, consider the given chemical properties and reactions. A white solid X decomposes upon heating to produce a white solid Y and an acidic gas Z. The same gas Z is released when X reacts with an acid. Additionally, Y can form when caustic soda (sodium hydroxide, NaOH) is exposed to the air.
02

Consider the reaction of substance X with an acid

The reaction of substance X with an acid releasing an acidic gas suggests it is a carbonate or a bicarbonate since both react with acids to liberate carbon dioxide (CO2), which is an acidic gas.
03

Analyze the thermal decomposition of compound X

Upon heating, substance X yields substance Y and gas Z. This also indicates that X is likely to be a bicarbonate (such as NaHCO3), which decomposes upon heating to form a carbonate (such as Na2CO3) and releases CO2 gas.
04

Evaluate the formation of substance Y from the exposure of caustic soda to the atmosphere

Caustic soda (NaOH) when left open in the atmosphere reacts with CO2 present in the air to form a white solid, which is sodium bicarbonate (NaHCO3). This rules out sodium carbonate (Na2CO3) as Y since it is not the direct product of the reaction between NaOH and CO2.
05

Selecting the correct option based on deductions

Based on the given clues and chemical logic, X is sodium bicarbonate (NaHCO3), Y is sodium carbonate (Na2CO3), and Z is carbon dioxide (CO2). The correct option that matches this description is (a).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Decomposition
Thermal decomposition is a chemical reaction where a single substance breaks down into two or more products due to the application of heat. Often observable in various salts and carbonates, this decomposition is crucial in understanding chemical stability and reactivity at different temperatures.

In our textbook exercise, the white solid 'X' undergoes thermal decomposition when heated, resulting in a new white solid 'Y' and an acidic gas 'Z'. This process is indicative of the substance 'X' being a bicarbonate. For example, sodium bicarbonate (\textbf{NaHCO}\(_3\)) decomposes to form sodium carbonate (\textbf{Na}\(_2\)\textbf{CO}\(_3\)) and carbon dioxide (\textbf{CO}\(_2\)) when heated:
\begin{align*}\textbf{2 NaHCO}_3 (s) \rightarrow \textbf{Na}_2\textbf{CO}_3 (s) + \textbf{CO}_2 (g) + \textbf{H}_2\textbf{O} (g)\text{.}\text{The production of carbon dioxide is a common outcome of the thermal decomposition of carbonates and bicarbonates.}
  • Thermal decomposition requires energy input, typically in the form of heat.
  • The products of decomposition depend on the substance and the conditions, like temperature and pressure.
  • This process is an example of an endothermic reaction, one that absorbs energy from the surroundings.
Carbonates and Bicarbonates
Carbonates (\textbf{CO}\(_3^{2-}\)) and bicarbonates (also known as hydrogen carbonates, \textbf{HCO}\(_3^-\)), are compounds that contain carbonate ions. These are widely known for their role in geological formations and biological systems, as well as for their numerous industrial applications.

In the context of our exercise, we encounter both a carbonate and a bicarbonate as substances 'Y' and 'X', respectively. Bicarbonates serve as an intermediate in the thermal decomposition chain, breaking down into carbonates when heated. Moreover, bicarbonates readily react with acids to produce carbon dioxide, a characteristic leveraged to identify compound 'X'.
  • Both carbonates and bicarbonates react with acids in acid-base reactions, often liberating \textbf{CO}\(_2\).
  • While carbonates are stable under standard conditions, bicarbonates decompose upon heating—a property used to distinguish between the two.
  • Bicarbonates, such as sodium bicarbonate (\textbf{NaHCO}\(_3\)), have various household and commercial uses, including baking and cleaning.
Acid-Base Reactions
Acid-base reactions are fundamental chemical processes where an acid and a base react to form a salt and typically water. In this context, 'acid' refers to a substance that can donate a proton, and 'base' refers to a substance that can accept a proton.

In our exercise, compound 'X' reacts with acid to produce gas 'Z'. This reaction signifies that 'X' could be a carbonate or a bicarbonate, both of which are common bases that interact with acids to create carbon dioxide (\textbf{CO}\(_2\)) and water (\textbf{H}\(_2\textbf{O}\)). The acid-base reaction of sodium bicarbonate with hydrochloric acid is a clear example:
\begin{align*}\textbf{NaHCO}_3 (s) + \textbf{HCl} (aq) \rightarrow \textbf{NaCl} (aq) + \textbf{CO}_2 (g) + \textbf{H}_2\textbf{O} (l)\text{.}
  • Acid-base reactions are essential in many biological systems and environmental processes.
  • The pH scale, which measures the acidity or basicity of a solution, is a crucial concept in understanding these reactions.
  • The ability to predict the outcomes of acid-base reactions helps chemists design processes that shape our daily lives, from digestion to industrial manufacturing.

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Most popular questions from this chapter

The ionisation energy of alkali metals decreases from \(L\) i to \(C s\) because (a) the atomic size increases from Li to Cs (b) the distance between nucleus and outermost orbital decreases from Li to Cs (c) dectropositive character decreases down the group (d) melting point decreases from Li to Cs.

Arrange the following elements in the order of the increasing electropositive character. Li, \(\mathrm{Na}, \mathrm{K}_{1} \mathrm{Rb}, \mathrm{Cs}\) (a) \(\mathrm{Li}>\mathrm{Na}>\mathrm{K}>\mathrm{Rb}>\mathrm{Cs}\) (b) \(\quad \mathrm{Li}<\mathrm{Na}<\mathrm{K}<\mathrm{Rb}<\mathrm{Cs}\) (c) \(\mathrm{Li}>\mathrm{Na}<\mathbb{K}<\mathrm{Rb}<\mathrm{Cs}\) (d) \(\mathrm{Na}>\mathrm{Li}>\mathrm{K}<\mathrm{Rb}<\mathrm{Cs}\)

Superoxides of alkali metals act as oxidising agents while normal oxides are basic in nature. The oxide which is paramagnetic in nature due to presence of unpaired electron is (a) \(\mathrm{Na}_{2} \mathrm{O}_{2}\) (b) \(\mathrm{KO}_{2}\) (c) \(\mathrm{Na}_{2} \mathrm{O}\) (d) \(\mathrm{K}_{2} \mathrm{O}_{2}\)

Match the column I with column II and mark the appropriate choice. \begin{tabular}{|l|l|l|l|} \hline \multicolumn{2}{|c|} { Column I } & \multicolumn{2}{c|} { Columi II = } \\\ \hline (A) & \(\mathrm{Na}\) & (i) & Crimson red \\ \hline (B) & \(\mathrm{K}\) & (ii) & Yellow \\ \hline (C) & \(\mathrm{Sr}\) & (iii) & Apple green \\ \hline (D) & \(\mathrm{Ba}\) & (iv) & Violet \\ \hline \end{tabular} (a) (A) \(\rightarrow(i),(B) \rightarrow(i i),(C) \rightarrow(i i i),(D) \rightarrow\) (iv) (b) \((\mathrm{A}) \rightarrow(\mathrm{ii}),(\mathrm{B}) \rightarrow(\mathrm{iv}),(\mathrm{C}) \rightarrow(\mathrm{i}),(\mathrm{D}) \rightarrow\) (iii) (c) \((\mathrm{A}) \rightarrow(\mathrm{iv}),(\mathrm{B}) \rightarrow\) (iii), (C) \(\rightarrow\) (ii), (D) \(\rightarrow\) (i) (d) (A) \(\rightarrow\) (iii), (B) \(\rightarrow\) (iv), (C) \(\rightarrow\) (i), (D) \(\rightarrow\) (ii)

When sodium reacts with excess of oxygen, the oxidation number of oxygen changes from (a) 0 to-1 (b) 0 to \(-2\) (c) \(-1\) to \(-2\) (d) No change.

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