H2SO4 + Al → Al2(SO4)3 + S + H2O
The given chemical reaction is a representation of the reaction between sulfuric acid (H₂SO₄) and aluminum (Al). It produces aluminum sulfate (Al₂(SO₄)₃), sulfur (S), and water (H₂O).
To balance this reaction, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Let’s break it down step by step.
Write down the unbalanced equation: \[ \text{H}_2\text{SO}_4 + \text{Al} \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{S} + \text{H}_2\text{O} \]
Identify the number of atoms for each element on both sides.
Balancing the equation:
- Aluminum (Al) must be balanced first. On the right side, there are 2 aluminum atoms (Al₂(SO₄)₃), so we need 2 on the left.
- Sulfur (S) in H₂SO₄ is present in 3 molecules of Al₂(SO₄)₃, contributing 3 sulfur atoms, and produces 1 sulfur atom as a product, making a total of 4.
- Hydrogen (H) is in water, which needs to be produced accordingly.
The balanced equation will then read as follows: \[ 3 \text{H}_2\text{SO}_4 + 2 \text{Al} \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3 \text{S} + 6 \text{H}_2\text{O} \]
- Confirm the balance:
Left Side:
- H: 3(2) = 6
- S: 3
- O: 3(4) = 12
- Al: 2
Right Side:
- H: 6 (from 3 H₂O)
- S: 3 (from Al₂(SO₄)₃ and 3 S)
- O: 12 (from Al₂(SO₄)₃ and 3 H₂O)
- Al: 2
Both sides match, confirming the reaction is balanced.
Final Balanced Reaction: \[ 3 \text{H}_2\text{SO}_4 + 2 \text{Al} \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3 \text{S} + 6 \text{H}_2\text{O} \]
