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 _2(g)[/tex] as a product
B. [tex]O _2(g)[/tex] as a reactant
C. [tex]O ( g )[/tex] as a product
D. [tex]2 O ( g )[/tex] as a reactant
Answer (1)
Add the two chemical equations.
Cancel out any common terms on both sides of the equation.
Combine the oxygen terms.
Observe that O 2 ( g ) appears as a reactant in the final equation.
The final answer is O 2 ( g ) as a reactant.
Explanation
Understanding the Problem We are given two chemical equations:
C ( s ) + 2 1 O 2 ( g ) → CO ( g )
CO ( g ) + 2 1 O 2 ( g ) → C O 2 ( g )
Our goal is to determine how oxygen appears in the final chemical equation after combining these two equations.
Adding the Equations To find the final chemical equation, we add the two given equations together:
C ( s ) + 2 1 O 2 ( g ) + CO ( g ) + 2 1 O 2 ( g ) → CO ( g ) + C O 2 ( g )
Canceling Common Terms Now, we simplify the equation by canceling out any common terms on both sides. In this case, CO ( g ) appears on both sides, so we cancel it out:
C ( s ) + 2 1 O 2 ( g ) + 2 1 O 2 ( g ) → C O 2 ( g )
Combining Oxygen Terms Combine the oxygen terms:
2 1 O 2 ( g ) + 2 1 O 2 ( g ) = 1 O 2 ( g ) = O 2 ( g )
So the final equation is:
C ( s ) + O 2 ( g ) → C O 2 ( g )
Conclusion In the final equation, O 2 ( g ) appears on the left side of the arrow, which means it is a reactant.
Examples
In the process of burning fuel, oxygen from the air combines with the fuel (like carbon) to produce energy and combustion products (like carbon dioxide). This problem illustrates how to track the role of oxygen in a chemical reaction, which is crucial for understanding combustion, respiration, and many other chemical processes. By understanding these reactions, we can optimize fuel efficiency and reduce pollution.
