Problem 89
Question
Assuming that water, common reagents (acids, bases, salts), and simple laboratory equipment are available, give a practical method to prepare (a) \(\mathrm{MgCl}_{2}\) from \(\mathrm{MgCO}_{3}(\mathrm{s}) ;\) (b) \(\mathrm{NaAl}(\mathrm{OH})_{4}\) from \(\mathrm{Na}(\mathrm{s})\) and \(\mathrm{Al}(\mathrm{s}) ;\) and (c) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) from \(\mathrm{NaCl}(\mathrm{s}).\)
Step-by-Step Solution
Verified Answer
The synthesis of \( \mathrm{MgCl}_{2} \) from \( \mathrm{MgCO}_{3} \) involves an acid-base reaction with \( \mathrm{HCl} \). The synthesis of \( \mathrm{NaAl}(\mathrm{OH})_{4} \) from \( \mathrm{Na}(\mathrm{s}) \) and \( \mathrm{Al}(\mathrm{s}) \) involves two steps, where \( \mathrm{Na} \) is first reacted with water to produce \( \mathrm{NaOH} \), and then \( \mathrm{Al} \) is reacted with \( \mathrm{NaOH} \) and water. The synthesis of \( \mathrm{Na}_{2} \mathrm{SO}_{4} \) from \( \mathrm{NaCl} \) involves a series of reactions with \( \mathrm{H}_{2}\mathrm{SO}_{4} \) to create \( \mathrm{Na}_{2}\mathrm{SO}_{4} \) and \( \mathrm{HCl} \).
1Step 1: Preparation of MgCl2 from MgCO3
To prepare \( \mathrm{MgCl}_{2} \) from \( \mathrm{MgCO}_{3} \), an acid-base reaction needs to be performed. The mineral magnesite, \( \mathrm{MgCO}_{3} \), is heated in the presence of hydrochloric acid, \( \mathrm{HCl} \), to produce \( \mathrm{MgCl}_{2} \), \( \mathrm{CO}_{2} \) gas, and water. The reaction is as follows: \( \mathrm{MgCO}_{3}(\mathrm{s}) + 2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{MgCl}_{2}(\mathrm{aq}) + \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2}\mathrm{O}(\mathrm{l}) \)
2Step 2: Preparation of NaAl(OH)4 from Na(s) and Al(s)
To prepare \( \mathrm{NaAl}(\mathrm{OH})_{4} \) from \( \mathrm{Na}(\mathrm{s}) \) and \( \mathrm{Al}(\mathrm{s}) \), a reaction with sodium hydroxide \( \mathrm{NaOH} \) is necessary. Sodium (Na) is reacted with water to produce \( \mathrm{NaOH} \) and Hydrogen \( \mathrm{H}_{2} \). The aluminium is then reacted with \( \mathrm{NaOH} \) in the presence of water to produce \( \mathrm{NaAl}(\mathrm{OH})_{4} \) and \( \mathrm{H}_{2} \). The reactions are as follows: \( 2\mathrm{Na} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{NaOH} + \mathrm{H}_{2} \), and \( \mathrm{Al} + 4\mathrm{NaOH} + \mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{NaAl}(\mathrm{OH})_{4} + \mathrm{H}_{2} \)
3Step 3: Preparation of Na2SO4 from NaCl
To prepare \( \mathrm{Na}_{2} \mathrm{SO}_{4} \) from \( \mathrm{NaCl} \), a series of reactions needs to be performed. Firstly, \( \mathrm{NaCl} \) is reacted with concentrated \( \mathrm{H}_{2}\mathrm{SO}_{4} \) to form \( \mathrm{NaHSO}_{4} \). Subsequently, \( \mathrm{NaHSO}_{4} \) is reacted with \( \mathrm{NaCl} \) to form \( \mathrm{Na}_{2}\mathrm{SO}_{4} \) and \( \mathrm{HCl} \). The reaction is performed in two steps as follows: \( \mathrm{NaCl} + \mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow \mathrm{NaHSO}_{4} + \mathrm{HCl} \) , and \( \mathrm{NaHSO}_{4} + \mathrm{NaCl} \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{4} + \mathrm{HCl} \)
Key Concepts
Preparation of Magnesium ChloridePreparation of Sodium AluminatePreparation of Sodium Sulfate
Preparation of Magnesium Chloride
To synthesize magnesium chloride \( \text{MgCl}_2 \) from magnesium carbonate \( \text{MgCO}_3 \), you need an acid, specifically hydrochloric acid \( \text{HCl} \). This process involves a straightforward chemical reaction where magnesium carbonate is mixed with hydrochloric acid. This results in the formation of magnesium chloride, carbon dioxide gas \( \text{CO}_2 \), and water. Here's how you can understand the reaction:
The process is as simple as adding dilute hydrochloric acid to magnesium carbonate. You'll observe the formation of bubbles due to the release of carbon dioxide gas. This is a sign that the reaction is proceeding. Once completed, the solution contains dissolved \( \text{MgCl}_2 \), which can be further purified or crystallized for use. The key takeaway is understanding the basic acid-base reaction taking place and the gas evolution as a clue to reaction progress.
- Reactants involved: \( \text{MgCO}_3 \) (solid) and \( \text{HCl} \) (aqueous).
- Reaction equation: \( \text{MgCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MgCl}_2(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \).
The process is as simple as adding dilute hydrochloric acid to magnesium carbonate. You'll observe the formation of bubbles due to the release of carbon dioxide gas. This is a sign that the reaction is proceeding. Once completed, the solution contains dissolved \( \text{MgCl}_2 \), which can be further purified or crystallized for use. The key takeaway is understanding the basic acid-base reaction taking place and the gas evolution as a clue to reaction progress.
Preparation of Sodium Aluminate
Sodium aluminate \( \text{NaAl(OH)}_4 \) can be synthesized using metallic sodium \( \text{Na} \) and aluminum \( \text{Al} \) by first generating sodium hydroxide, \( \text{NaOH} \), and then reacting it further with aluminum in the presence of water. This might sound complicated, but let's simplify it.
The reaction between sodium and water is exothermic, releasing heat and hydrogen gas. Therefore, caution must be exercised. Ensure that this is done in a controlled environment. The subsequent reaction with aluminum involves the hydroxide ion \( \text{OH}^- \) participating in breaking down the aluminum into aluminate ions, which are incorporated into the sodium compound.
- Sodium reaction with water: \( 2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 \), where sodium reacts vigorously with water to form sodium hydroxide and hydrogen gas.
- Aluminum reaction with sodium hydroxide: Now, the \( \text{NaOH} \) formed is used to react with aluminum. The reaction \( \text{Al} + 4\text{NaOH} + \text{H}_2\text{O} \rightarrow \text{NaAl(OH)}_4 + \text{H}_2 \) leads to the formation of sodium aluminate.
The reaction between sodium and water is exothermic, releasing heat and hydrogen gas. Therefore, caution must be exercised. Ensure that this is done in a controlled environment. The subsequent reaction with aluminum involves the hydroxide ion \( \text{OH}^- \) participating in breaking down the aluminum into aluminate ions, which are incorporated into the sodium compound.
Preparation of Sodium Sulfate
Creating sodium sulfate, \( \text{Na}_2\text{SO}_4 \), from sodium chloride \( \text{NaCl} \) involves two main reaction steps with sulfuric acid, \( \text{H}_2\text{SO}_4 \). This is an important reaction series in the production of industrial chemicals.
This process showcases an interesting example of salt metathesis or double displacement reaction. The key point is the intermediate formation of sodium hydrogen sulfate which at a further stage produces sodium sulfate. Remember, handling \( \text{H}_2\text{SO}_4 \) requires care due to its corrosiveness, and the emitted \( \text{HCl} \) gas needs appropriate ventilation controls.
- Initial reaction with sulfuric acid: The first step is to react sodium chloride with \( \text{H}_2\text{SO}_4 \) to form sodium hydrogen sulfate and hydrochloric acid. The equation is \( \text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl} \).
- Formation of sodium sulfate: The second step is to take another molecule of sodium chloride and react it with the sodium hydrogen sulfate to yield \( \text{Na}_2\text{SO}_4 \) and additional hydrochloric acid. The equation is \( \text{NaHSO}_4 + \text{NaCl} \rightarrow \text{Na}_2\text{SO}_4 + \text{HCl} \).
This process showcases an interesting example of salt metathesis or double displacement reaction. The key point is the intermediate formation of sodium hydrogen sulfate which at a further stage produces sodium sulfate. Remember, handling \( \text{H}_2\text{SO}_4 \) requires care due to its corrosiveness, and the emitted \( \text{HCl} \) gas needs appropriate ventilation controls.
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