Let's address the problem step-by-step by analyzing the given intermediate chemical equations:
1. The first intermediate equation is:
[tex]\[
C (s) + \frac{1}{2} O_2 (g) \rightarrow CO (g)
\][/tex]
2. The second intermediate equation is:
[tex]\[
CO (g) + \frac{1}{2} O_2 (g) \rightarrow CO_2 (g)
\][/tex]
To form the final chemical equation, we need to combine the two reactions. Let's proceed with this:
Step 1: Write both equations together:
[tex]\[
C (s) + \frac{1}{2} O_2 (g) \rightarrow CO (g)
\][/tex]
[tex]\[
CO (g) + \frac{1}{2} O_2 (g) \rightarrow CO_2 (g)
\][/tex]
Step 2: Combine the equations to form a single equation:
[tex]\[
C (s) + \frac{1}{2} O_2 (g) + CO (g) + \frac{1}{2} O_2 (g) \rightarrow CO (g) + CO_2 (g)
\][/tex]
Step 3: Observe that CO appears as both a product in the first reaction and a reactant in the second reaction:
- [tex]\(CO (g)\)[/tex] on the right-hand side of the first equation
- [tex]\(CO (g)\)[/tex] on the left-hand side of the second equation
Since [tex]\(CO (g)\)[/tex] appears on both sides of the combined equation, it cancels out.
Step 4: Cancel [tex]\(CO (g)\)[/tex] from both sides and simplify:
[tex]\[
C (s) + \frac{1}{2} O_2 (g) + \frac{1}{2} O_2 (g) \rightarrow CO_2 (g)
\][/tex]
Combining the oxygen terms:
[tex]\[
C (s) + O_2 (g) \rightarrow CO_2 (g)
\][/tex]
Final Decision:
By examining the entire process, it's clear that CO is an intermediary that gets consumed and produced in equal amounts during the reaction pathway. Therefore, it should be canceled out in the final chemical equation.
So, the final decision regarding CO in forming the final chemical equation is:
Cancel out [tex]\(CO\)[/tex] because it appears as a reactant in one intermediate reaction and a product in the other intermediate reaction.
