[VISUAL] 9. A mixture of CO(g) and O2(g) is placed in a container, as shown above. A reaction occurs — Stoichiometry Chemistry Question

STEPS:

1. Write and balance the chemical equation for the reaction:
Carbon monoxide ($\text{CO}$) reacts with oxygen gas ($\text{O}_2$) to produce carbon dioxide ($\text{CO}_2$). The balanced chemical equation is:
$$2\ \text{CO}(g) + \text{O}_2(g) \rightarrow 2\ \text{CO}_2(g)$$
This stoichiometry tells us that 2 molecules of $\text{CO}$ react with exactly 1 molecule of $\text{O}_2$ to produce 2 molecules of $\text{CO}_2$.

2. Count the initial reactant molecules in the starting container:
Using the key provided (where gray circles represent $\text{C}$ and white circles represent $\text{O}$):
* $\text{CO}$ molecules (one gray circle bonded to one white circle): Count them systematically to find there are exactly 6 $\text{CO}$ molecules.
* $\text{O}_2$ molecules (two bonded white circles): Count them systematically to find there are exactly 4 $\text{O}_2$ molecules.

3. Identify the limiting reactant:
Determine how much $\text{O}_2$ is needed to react completely with all 6 molecules of $\text{CO}$:
$$6\text{ molecules of CO} \times \frac{1\text{ molecule of }\text{O}_2}{2\text{ molecules of CO}} = \mathbf{3\text{ molecules of }\text{O}_2\text{ needed}}$$
Because we have 4 molecules of $\text{O}_2$ initially and only need 3:
* $\text{CO}$ is the limiting reactant (it will be completely consumed).
* $\text{O}_2$ is the excess reactant.

4. Calculate the contents of the container after the reaction goes to completion:
* $\text{CO}$ remaining: Since it is the limiting reactant, $6 - 6 = \mathbf{0\text{ molecules}}$ remain.
* $\text{CO}_2$ produced: Since $\text{CO}$ and $\text{CO}_2$ have a $1:1$ stoichiometric ratio ($2:2$), completely consuming 6 molecules of $\text{CO}$ produces exactly 6 molecules of $\text{CO}_2$ (each represented by one central gray circle bonded to two white circles).
* $\text{O}_2$ remaining: Out of the initial 4 molecules, 3 are consumed, leaving:
$$4\text{ initial} - 3\text{ consumed} = \mathbf{1\text{ molecule of }\text{O}_2\text{ remaining}}$$

5. Select the corresponding box:
The final container must contain exactly 6 $\text{CO}_2$ molecules and 1 unreacted $\text{O}_2$ molecule. Looking at the choices, Box B matches this composition perfectly, making Option B the correct answer.

*

WHY_OTHERS_WRONG:

• Option A is incorrect (Box A): This box contains 6 $\text{CO}_2$ molecules but 0 $\text{O}_2$ molecules. This representation violates the law of conservation of mass because the two oxygen atoms belonging to the excess, unreacted $\text{O}_2$ molecule have completely vanished from the system.
• Option C is incorrect (Box C): This box represents the starting system with no reaction having occurred (it still contains 6 $\text{CO}$ molecules and 4 $\text{O}_2$ molecules). This contradicts the prompt, which states that a reaction occurs and proceeds as completely as possible.
• Option D is incorrect (Box D): This box shows 6 $\text{CO}_2$ molecules along with 4 $\text{O}_2$ molecules. For this composition to exist, the final system would need to contain 14 oxygen atoms and 6 carbon atoms, which is far more than the initial starting atoms.