To find the final chemical equation from the given intermediate chemical equations, let's analyze each step systematically.
We are given two intermediate equations:
1. [tex]\( P_4O_6(s) \rightarrow P_4(s) + 3O_2(g) \)[/tex]
2. [tex]\( P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s) \)[/tex]
The goal is to derive the single combined chemical equation.
Step-by-Step Solution:
1. Identify the species involved in each step:
- For the first intermediate equation:
- Reactant: [tex]\( P_4O_6(s) \)[/tex]
- Products: [tex]\( P_4(s) \)[/tex], [tex]\( O_2(g) \)[/tex]
- For the second intermediate equation:
- Reactants: [tex]\( P_4(s) \)[/tex], [tex]\( O_2(g) \)[/tex]
- Product: [tex]\( P_4O_{10}(s) \)[/tex]
2. Combine the two reactions:
- The product [tex]\( P_4(s) \)[/tex] formed in the first reaction serves as the reactant in the second reaction. Therefore, we can combine the two reactions by ensuring the [tex]\( P_4(s) \)[/tex] produced is used completely.
[tex]\[ P_4O_6(s) \rightarrow P_4(s) + 3O_2(g) \][/tex]
[tex]\[ P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s) \][/tex]
3. Sum the reactions:
- By adding the two reactions together, we get:
[tex]\[ P_4O_6(s) \rightarrow P_4(s) + 3O_2(g) \][/tex]
[tex]\[ P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s) \][/tex]
When we add these reactions, the [tex]\( P_4(s) \)[/tex] cancels out from both sides:
[tex]\[ P_4O_6(s) + 5O_2(g) \rightarrow P_4(s) + 3O_2(g) + P_4O_{10}(s) \][/tex]
Simplifying further:
[tex]\[ P_4O_6(s) + (5 - 3)O_2(g) \rightarrow P_4O_{10}(s) \][/tex]
[tex]\[ P_4O_6(s) + 2O_2(g) \rightarrow P_4O_{10}(s) \][/tex]
Final Chemical Equation:
Therefore, the final chemical equation derived from the intermediate steps is:
[tex]\[ P_4O_6(s) + 2O_2(g) \rightarrow P_4O_{10}(s) \][/tex]
This is the correct and final chemical equation resulting from combining the intermediate steps given.
