In the quantitative oxidation of Hydrogen sulphide () to elemental iodine () using Potassium iodide — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
In the quantitative oxidation of Hydrogen sulphide ($H_2S$) to elemental iodine ($I_2$) using Potassium iodide ($KI$) and Sulphuric acid ($H_2SO_4$), the balanced reaction is: $8KI + 5H_2SO_4 \rightarrow 4K_2SO_4 + 4I_2 + H_2S + 4H_2O$. What exact volume (in Litres) of a $0.20\text{ M } H_2SO_4$ standard solution is required to synthesize exactly $34.0\text{ g}$ of $H_2S$ gas?
💡 Solution & Explanation
Step 1: Calculate the moles of the target product ($H_2S$). Molar mass of $H_2S = 2 + 32 = 34\text{ g/mol}$. Moles of $H_2S = \frac{34.0\text{ g}}{34\text{ g/mol}} = 1.0\text{ mole}$. Step 2: Utilize the stoichiometric ratio from the balanced equation. The equation dictates that $5\text{ moles}$ of $H_2SO_4$ are required to yield exactly $1\text{ mole}$ of $H_2S$. Moles of $H_2SO_4$ required = $5.0\text{ moles}$. Step 3: Determine the volume of the acid solution. Volume (L) = $\frac{\text{Moles}}{\text{Molarity}} = \frac{5.0}{0.20} = 25\text{ Litres}$.