Step 1: Understanding the unbalanced chemical equation:
The given unbalanced chemical equation is:
\(p \, \text{MnO}_2 + q \, \text{Al} \rightarrow r \, \text{Mn} + s \, \text{Al}_2\text{O}_3\)
In this equation, manganese dioxide (MnO₂) reacts with aluminum (Al) to form manganese (Mn) and aluminum oxide (Al₂O₃). We need to balance the number of atoms of each element on both sides of the equation.
Step 2: Balancing the equation step by step:
1. Balance the manganese (Mn):
On the left-hand side, manganese is present as part of MnO₂. On the right-hand side, manganese is in its elemental form (Mn). To balance manganese, we put a coefficient of 3 in front of MnO₂ and a coefficient of 3 in front of Mn on the right-hand side:
\(3 \, \text{MnO}_2 + q \, \text{Al} \rightarrow 3 \, \text{Mn} + s \, \text{Al}_2\text{O}_3\)
2. Balance the oxygen (O):
On the left-hand side, there are 6 oxygen atoms in 3 MnO₂ molecules. On the right-hand side, oxygen is part of Al₂O₃. Since Al₂O₃ contains 3 oxygen atoms per molecule, we need 2 Al₂O₃ molecules to balance the oxygen atoms:
\(3 \, \text{MnO}_2 + q \, \text{Al} \rightarrow 3 \, \text{Mn} + 2 \, \text{Al}_2\text{O}_3\)
3. Balance the aluminum (Al):
On the right-hand side, there are 4 aluminum atoms in 2 Al₂O₃ molecules (since each Al₂O₃ contains 2 aluminum atoms). Therefore, we need 4 aluminum atoms on the left-hand side, so we put a coefficient of 4 in front of Al:
\(3 \, \text{MnO}_2 + 4 \, \text{Al} \rightarrow 3 \, \text{Mn} + 2 \, \text{Al}_2\text{O}_3\)
4. Verify the final balance of atoms:
- Manganese: 3 atoms on both sides (3 MnO₂ → 3 Mn)
- Oxygen: 6 atoms on both sides (3 MnO₂ → 2 Al₂O₃)
- Aluminum: 4 atoms on both sides (4 Al → 2 Al₂O₃)
Step 3: Conclusion:
The balanced chemical equation is:
\(3 \, \text{MnO}_2 + 4 \, \text{Al} \rightarrow 3 \, \text{Mn} + 2 \, \text{Al}_2\text{O}_3\)
Thus, the values of the coefficients \(p\), \(q\), \(r\), and \(s\) are 3, 4, 3, 2, respectively.