Problem 36
Question
Aluminium reacts with concentrated \(\mathrm{HCl}\) and concentrated \(\mathrm{NaOH}\) to liberate the gases \(\ldots \ldots\) respectively. (a) \(\mathrm{H}_{2}\) and \(\mathrm{H}_{2}\) (b) \(\mathrm{O}_{2}\) and \(\mathrm{O}_{2}\) (c) \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2}\) (d) \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\)
Step-by-Step Solution
Verified Answer
(a) \(\mathrm{H}_{2}\) and \(\mathrm{H}_{2}\)
1Step 1: Identify the reactions
Understand that aluminum reacts with both hydrochloric acid (HCl) and sodium hydroxide (NaOH) to release gases. We need to determine these gases for each reaction.
2Step 2: Reaction with Hydrochloric Acid
When aluminum (\( \mathrm{Al} \)) reacts with hydrochloric acid (\( \mathrm{HCl} \)), hydrogen gas (\( \mathrm{H_2} \)) is released. The balanced chemical equation for this reaction is: \[ 2\mathrm{Al} + 6\mathrm{HCl} \rightarrow 2\mathrm{AlCl_3} + 3\mathrm{H_2} \]. The gas released from this reaction is \( \mathrm{H_2} \).
3Step 3: Reaction with Sodium Hydroxide
When aluminum reacts with sodium hydroxide (\( \mathrm{NaOH} \)), hydrogen gas (\( \mathrm{H_2} \)) is also released. The balanced chemical equation is: \[ 2\mathrm{Al} + 2\mathrm{NaOH} + 6\mathrm{H_2O} \rightarrow 2\mathrm{NaAl(OH)_4} + 3\mathrm{H_2} \]. The gas released from this reaction is \( \mathrm{H_2} \).
4Step 4: Conclusion
The gases released from both reactions are hydrogen (\( \mathrm{H_2} \)) and hydrogen (\( \mathrm{H_2} \)). Therefore, the correct choice is option (a).
Key Concepts
Hydrochloric acid reactionSodium hydroxide reactionHydrogen gas production
Hydrochloric acid reaction
When aluminum (2Al1) interacts with hydrochloric acid (2HCl1), an interesting chemical reaction occurs. This reaction results in the liberation of hydrogen gas (2H_21). The process begins as the aluminum displaces hydrogen from the hydrochloric acid. This happens because aluminum is a more reactive metal than hydrogen. The activity series of metals plays a crucial role in predicting this displacement reaction.
To break it down further, in the reaction between aluminum and hydrochloric acid:
To break it down further, in the reaction between aluminum and hydrochloric acid:
- Aluminum goes from a metal to forming aluminum chloride (2AlCl_31).
- Hydrogen ions from the acid are reduced to form hydrogen gas.
Sodium hydroxide reaction
Aluminum also reacts with sodium hydroxide (2NaOH1) to produce hydrogen gas. This might be surprising to some, as 2NaOH1 is a strong base rather than an acid like 2HCl1. However, aluminum can react with 2NaOH1 in the presence of water to liberate hydrogen gas due to its amphoteric nature.
Aluminum's amphoteric property means it can react with both acids and bases. During the reaction with sodium hydroxide:
Aluminum's amphoteric property means it can react with both acids and bases. During the reaction with sodium hydroxide:
- Aluminum forms sodium aluminate (2NaAl(OH)_41).
- In this process, water molecules also play an essential role to release hydrogen gas.
Hydrogen gas production
Hydrogen gas production is a key result of aluminum's interaction with both hydrochloric acid and sodium hydroxide. In each of these reactions, hydrogen gas is consistently formed, albeit through slightly different pathways.
This production of hydrogen is of interest for a variety of scientific and industrial reasons:
This production of hydrogen is of interest for a variety of scientific and industrial reasons:
- Hydrogen gas is a clean fuel, making its production important for energy applications.
- Knowing how to generate hydrogen via chemical reactions can aid in understanding and developing alternative energy resources.
- Additionally, learning about these reactions helps students understand core principles of chemistry, such as reactivity and the concept of amphoterism.
Other exercises in this chapter
Problem 34
In which of the following reactions does hydrogen peroxide acts as a reducing agent? (a) \(2 \mathrm{KI}+\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{SO
View solution Problem 35
In which of the following reactions does \(\mathrm{H}_{2} \mathrm{O}_{2}\) acts as a reducing agent? (a) \(\mathrm{PbO}_{2}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{
View solution Problem 38
For decolourization of one mole of \(\mathrm{KMnO}_{4}\) the number of moles of \(\mathrm{H}_{2} \mathrm{O}_{2}\) required are (a) \(1 / 2\) (b) \(3 / 2\) (c) \
View solution Problem 39
A commercial sample of hydrogen peroxide is labelled as 10 volume, its percentage strength is (a) \(1 \%\) (b) \(3 \%\) (c) \(10 \%\) (d) \(90 \%\)
View solution