Consider the following intermediate chemical equations:

[tex]\[
\begin{array}{l}
C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g) \\
CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g)
\end{array}
\][/tex]

How will oxygen appear in the final chemical equation?

A. [tex]\( O(g) \)[/tex] as a product
B. [tex]\( O_2(g) \)[/tex] as a reactant
C. [tex]\( O(g) \)[/tex] as a product
D. [tex]\( O_2(g) \)[/tex] as a reactant



Answer :

To determine how oxygen appears in the final chemical equation derived from the given intermediate chemical equations, let’s combine the two equations step by step. Here are the given intermediate equations:

1. [tex]\( C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g) \)[/tex]
2. [tex]\( CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \)[/tex]

Step-by-step combination:

1. Add the two reactions:
[tex]\[ \begin{aligned} &\text{First Equation:} & C(s) + \frac{1}{2} O_2(g) &\rightarrow CO(g) \\ &\text{Second Equation:} & CO(g) + \frac{1}{2} O_2(g) &\rightarrow CO_2(g) \\ \end{aligned} \][/tex]

2. When we add these equations together, we combine the reactants and the products:
[tex]\[ C(s) + \frac{1}{2} O_2(g) + CO(g) + \frac{1}{2} O_2(g) \rightarrow CO(g) + CO_2(g) \][/tex]

3. Simplify the equation by canceling out the [tex]\( CO(g) \)[/tex] on both the reactant and product sides:
[tex]\[ C(s) + \frac{1}{2} O_2(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \][/tex]

4. Combine the [tex]\( O_2(g) \)[/tex] terms on the left side:
[tex]\[ C(s) + 1 O_2(g) \rightarrow CO_2(g) \][/tex]

After combining and simplifying the two intermediate reactions, the overall reaction becomes:
[tex]\[ C(s) + O_2(g) \rightarrow CO_2(g) \][/tex]

From this final combined equation, it is evident that:

- [tex]\( O_2(g) \)[/tex] appears as a reactant.

Therefore, the correct answer regarding how oxygen appears in the final chemical equation is:
[tex]\[ O_2(g) \text{ as a reactant} \][/tex]