A mixture of H2(g) and O2(g) is placed in a container as represented above. [VISUAL] The H2(g) and O — Stoichiometry Chemistry Question

STEPS:

1. Count the starting reactant molecules in the initial mixture:
By examining the particulate diagram of the reactants, there are exactly four molecules of $\text{H}_2$ (represented as pairs of dark spheres) and four molecules of $\text{O}_2$ (represented as pairs of unshaded spheres).

2. Write and balance the chemical equation:
The reaction of hydrogen gas and oxygen gas to form water vapor is represented by the balanced equation:
$$2\ \text{H}_2(g) + \text{O}_2(g) \rightarrow 2\ \text{H}_2\text{O}(g)$$
This stoichiometric relationship dictates that two molecules of $\text{H}_2$ react with one molecule of $\text{O}_2$ to produce two molecules of $\text{H}_2\text{O}$.

3. Identify the limiting reactant:
* Using the $2:1$ stoichiometric ratio, the four available molecules of $\text{H}_2$ require exactly:
$$4\text{ molecules } \text{H}_2 \times \frac{1\text{ molecule } \text{O}_2}{2\text{ molecules } \text{H}_2} = \mathbf{2\text{ molecules of } \text{O}_2}$$
* Because we have four molecules of $\text{O}_2$ in the container, but only two are needed to react, oxygen is in excess.
* Consequently, $\text{H}_2$ is the limiting reactant and will be completely consumed.

4. Calculate the contents of the container after the reaction goes to completion:
* Water molecules produced: The four molecules of limiting $\text{H}_2$ will react completely to form exactly four molecules of $\text{H}_2\text{O}$.
* Excess oxygen molecules remaining: Two of the four original $\text{O}_2$ molecules are consumed in the reaction, leaving exactly two molecules of unreacted $\text{O}_2$ remaining in the vessel.
* Hydrogen molecules remaining: Zero molecules of $\text{H}_2$ remain since it was the limiting reactant.

5. Identify the correct diagram:
The particulate diagram representing completion must contain exactly four $\text{H}_2\text{O}$ molecules (each shown as one unshaded oxygen sphere bonded to two dark hydrogen spheres) and two intact $\text{O}_2$ molecules (each shown as a pair of unshaded oxygen spheres). This matches the arrangement in Diagram B (Option B).

*

WHY_OTHERS_WRONG:

• Option A is incorrect: This diagram represents only the four produced water molecules but fails to show any remaining oxygen. This directly violates the Law of Conservation of Mass because four oxygen atoms from the reactant side are missing.
• Option C is incorrect: Although the absolute atom count is stoichiometric, oxygen is represented as four individual, isolated O atoms instead of two diatomic $\text{O}_2$ molecules. Gaseous oxygen exists as diatomic $\text{O}_2$, and the reaction conditions do not favor molecular dissociation (if conditions allowed $\text{O}_2$ to dissociate, the $\text{H}_2\text{O}$ molecules would be dissociated as well).
• Option D is incorrect: This diagram shows eight $\text{H}_2\text{O}$ molecules and two $\text{O}_2$ molecules. This arrangement violates the Law of Conservation of Mass by displaying eight extra hydrogen atoms and four extra oxygen atoms that were not present in the starting mixture.