A highly corrosive mixture containing exactly of formic acid () and of oxalic acid () is treated wit — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
A highly corrosive mixture containing exactly $2.3\text{ g}$ of formic acid ($HCOOH$) and $4.5\text{ g}$ of oxalic acid ($H_2C_2O_4$) is treated with concentrated sulfuric acid ($H_2SO_4$). The entirely evolved gaseous mixture is subsequently passed through solid Potassium Hydroxide ($KOH$) pellets. Which of the following statements accurately describe this multi-step analytical process?
💡 Solution & Explanation
Step 1: Dehydration reactions. $H_2SO_4$ dehydrates both acids (Statement A is True). $HCOOH \rightarrow H_2O + CO$. Moles = $2.3/46 = 0.05\text{ mol}$. Produces $0.05\text{ mol}$ of $CO$. $H_2C_2O_4 \rightarrow H_2O + CO + CO_2$. Moles = $4.5/90 = 0.05\text{ mol}$. Produces $0.05\text{ mol}$ of $CO$ and $0.05\text{ mol}$ of $CO_2$. Step 2: Total $CO$ produced = $0.05 + 0.05 = 0.1\text{ moles}$ (Statement C is True). Step 3: Absorption. $KOH$ is a strong base and selectively absorbs the acidic $CO_2$ gas, not the neutral $CO$ gas (Statement B is False). Step 4: Remaining gas is purely $CO$ ($0.1\text{ moles}$). Mass of $CO$ = $0.1 \times 28 = 2.8\text{ g}$ (Statement D is True).