Exactly of an unknown active metal reacts completely with halogens to form exactly of its correspond — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
Exactly $5.6\text{ g}$ of an unknown active metal reacts completely with halogens to form exactly $12.7\text{ g}$ of its corresponding metal chloride salt. What is the precise equivalent weight of this unknown metal?
💡 Solution & Explanation
Step 1: Establish the mass relationship. Mass of the pure metal ($W_M$) = $5.6\text{ g}$. Mass of the metal chloride ($W_{MCl}$) = $12.7\text{ g}$. Therefore, mass of the combined Chlorine ($W_{Cl}$) = $12.7 - 5.6 = 7.1\text{ g}$. Step 2: Apply the fundamental definition of Equivalent Weight. The equivalent weight of a metal is the mass of the metal that chemically combines with exactly one equivalent mass ($35.5\text{ g}$) of Chlorine. Step 3: Calculate the equivalent weight ($E$). $E = \left(\frac{W_M}{W_{Cl}}\right) \times 35.5 = \left(\frac{5.6}{7.1}\right) \times 35.5$. $E = 0.7887 \times 35.5 = 28$. Alternatively, using Equivalents: $5.6/E = 12.7/(E + 35.5) \Rightarrow E = 28$.