0.50 g of a solid mixture containing and requires exactly 30 mL of a 0.25 N solution for complete ne — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

Let mass of $K_2CO_3$ (MW=138, E=69) be $x$. Mass of $Li_2CO_3$ (MW=74, E=37) is $(0.5 - x)$. Equivalents of acid used = $30 \times 0.25 / 1000 = 0.0075$ eq. By Law of Equivalence: $\frac{x}{69} + \frac{0.5 - x}{37} = 0.0075$. Multiplying by $69 \times 37 = 2553$ gives: $37x + 69(0.5 - x) = 0.0075 \times 2553 \Rightarrow 37x + 34.5 - 69x = 19.1475 \Rightarrow -32x = -15.3525 \Rightarrow x \approx 0.48$ g. Mass percentage = $(0.48 / 0.5) \times 100 = 96\%$.