Reaction 1: CaC2(s) + 2 H2O(l) ⇄ C2H2(g) + Ca(OH)2(s) Reaction 2: NaOCl(aq) + 2 HCl(aq) ⇄ Cl2(g) + N — Stoichiometry Chemistry Question

STEPS:

1. Analyze the stoichiometry of Reaction 2:
Look at the balanced chemical equation representing the reaction:
$$\text{NaOCl}(aq) + 2\ \text{HCl}(aq) \rightarrow \text{Cl}_2(g) + \text{NaCl}(aq) + \text{H}_2\text{O}(l)$$
According to the coefficients, the stoichiometric mole ratio between the reactant of interest ($\text{NaOCl}$) and the gas product ($\text{Cl}_2$) is $1:1$.
2. Calculate the moles of $\text{NaOCl}$ that reacted:
Since the reaction goes to completion and produces $0.010\text{ mol}$ of $\text{Cl}_2(g)$, we can use the 1:1 stoichiometric ratio to determine that the amount of $\text{NaOCl}$ initially present must also be exactly:
$$0.010\text{ mol Cl}_2 \times \frac{1\text{ mol NaOCl}}{1\text{ mol Cl}_2} = \mathbf{0.010\text{ mol NaOCl}}\ \text{}$$
3. Convert the volume of the $\text{NaOCl}$ solution to liters:
The initial volume of the $\text{NaOCl}$ solution is given as $100.\text{ mL}$. Convert this volume to liters so that we can calculate molarity in standard units ($\text{mol/L}$):
$$100.\text{ mL} \times \frac{1\text{ L}}{1000\text{ mL}} = \mathbf{0.100\text{ L}}$$
4. Calculate the molarity of the $\text{NaOCl}$ solution:
Molarity ($M$) is defined as the number of moles of solute divided by the total volume of the solution in liters:
$$\text{Molarity } (M) = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{0.010\text{ mol NaOCl}}{0.100\text{ L solution}} = \mathbf{0.10\text{ M}}\ \text{}$$
This calculation shows that the starting concentration of $\text{NaOCl}$ was $0.10\text{ M}$, making Option C the correct answer.

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WHY_OTHERS_WRONG:

• Option A is incorrect (0.0010 M): This value is off by two orders of magnitude. A student might arrive at this number if they made a decimal placement error during calculation or incorrectly converted $100.\text{ mL}$ to $10.\text{ L}$ instead of $0.100\text{ L}$.
• Option B is incorrect (0.010 M): This value is numerically equal to the number of moles of solute ($0.010\text{ mol}$). This represents a common mistake where a student forgets to divide the moles by the volume of the solution in liters, effectively assuming the starting solution volume was exactly $1.0\text{ L}$.
• Option D is incorrect (1.0 M): This value is off by a factor of 10. A student might get this answer if they incorrectly multiplied the moles by the volume in liters rather than dividing, or if they divided $0.010\text{ mol}$ by $0.010\text{ L}$ (which is only $10.\text{ mL}$) instead of $0.100\text{ L}$.