MnO4 − + 5 Fe2+ + 8 H+ ➔ Mn2+ + 5 Fe3+ + 4 H2O In the reaction represented above, the number of MnO4 — Stoichiometry Chemistry Question

STEPS:

1. Analyze the balanced chemical equation: The given equation represents a redox reaction in acidic solution:
$$\text{MnO}_4^-(aq) + 5\text{ Fe}^{2+}(aq) + 8\text{ H}^+(aq) \rightarrow \text{Mn}^{2+}(aq) + 5\text{ Fe}^{3+}(aq) + 4\text{ H}_2\text{O}(l)$$
2. Identify the stoichiometric coefficients of the species of interest: Locate the coefficients for each reactant and product:
* Permanganate ion ($\text{MnO}_4^-$): $1$
* Iron(II) ion ($\text{Fe}^{2+}$): $5$
* Hydrogen ion ($\text{H}^+$): $8$
* Iron(III) ion ($\text{Fe}^{3+}$): $5$
* Water ($\text{H}_2\text{O}$): $4$
3. Set up the mole-to-mole ratios: Use the coefficients to write stoichiometric conversion factors between the reacting $\text{MnO}_4^-$ ions and the other species:
* With $\text{Fe}^{2+}$: $\frac{1\text{ MnO}_4^-}{5\text{ Fe}^{2+}}$ (For every 5 $\text{Fe}^{2+}$ ions consumed, 1 $\text{MnO}_4^-$ ion reacts)
* With $\text{H}^+$: $\frac{1\text{ MnO}_4^-}{8\text{ H}^+}$ (For every 8 $\text{H}^+$ ions consumed, 1 $\text{MnO}_4^-$ ion reacts)
* With $\text{Fe}^{3+}$: $\frac{1\text{ MnO}_4^-}{5\text{ Fe}^{3+}}$ (For every 5 $\text{Fe}^{3+}$ ions produced, 1 $\text{MnO}_4^-$ ion reacts)
* With $\text{H}_2\text{O}$: $\frac{1\text{ MnO}_4^-}{4\text{ H}_2\text{O}}$ (For every 4 $\text{H}_2\text{O}$ molecules produced, 1 $\text{MnO}_4^-$ ion reacts)
4. Relate the quantities algebraically:
* $\text{Number of }\text{MnO}_4^-\text{ ions} = \frac{1}{5} \times (\text{Number of }\text{Fe}^{2+}\text{ ions consumed})$
* This matches Option A exactly.

*

WHY_OTHERS_WRONG:

• B is incorrect: The stoichiometric ratio between $\text{MnO}_4^-$ and $\text{H}^+$ is 1:8. Therefore, the number of reacting $\text{MnO}_4^-$ ions is one-eighth ($\frac{1}{8}$) the number of $\text{H}^+$ ions consumed, not eight times as many.
• C is incorrect: The stoichiometric ratio between $\text{MnO}_4^-$ and $\text{Fe}^{3+}$ is 1:5. Therefore, the number of reacting $\text{MnO}_4^-$ ions is one-fifth ($\frac{1}{5}$) the number of $\text{Fe}^{3+}$ ions produced, not five times as many.
• D is incorrect: The stoichiometric ratio between $\text{MnO}_4^-$ and $\text{H}_2\text{O}$ is 1:4. Therefore, the number of reacting $\text{MnO}_4^-$ ions is one-fourth ($\frac{1}{4}$) the number of $\text{H}_2\text{O}$ molecules produced, not one-half.