This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Welcome back Gobte — missed you this week. a) Relative atomic mass is the weighted average mass of an atom of an element, taking into account the masses and relative abundances of all its naturally occurring isotopes, compared to (1)/(12) the mass of a carbon-12 atom. b) Step 1: Write the balanced chemical equation. The reaction between a metal carbonate X₂CO₃ and hydrochloric acid (HCl) produces a metal chloride (XCl), water, and carbon dioxide. X_2CO_3(s) + 2HCl(aq) → 2XCl(aq) + H_2O(l) + CO_2(g) Step 2: Calculate the number of moles of hydrochloric acid (HCl). Given: Volume of HCl = 23.6 cm^3 = 0.0236 dm^3 Concentration of HCl = 0.15 M Moles of HCl = Concentration × Volume Moles of HCl = 0.15 mol/dm^3 × 0.0236 dm^3 Moles of HCl = 0.00354 mol Step 3: Calculate the number of moles of the carbonate (X₂CO₃) used. From the balanced equation, the mole ratio of X₂CO₃ to HCl is 1:2. Moles of X_2CO_3 = (1)/(2) × Moles of HCl Moles of X_2CO_3 = (1)/(2) × 0.00354 mol Moles of X_2CO_3 = 0.00177 mol Step 4: Determine the Relative molecular mass of X₂CO₃. Given: Mass of X₂CO₃ = 0.245 g Moles of X₂CO₃ = 0.00177 mol Relative molecular mass (M_r) = MassMoles M_r of X_2CO_3 = 0.245 g0.00177 mol M_r of X_2CO_3 ≈ 138.42 g/mol Step 5: Calculate the Relative Atomic Mass (RAM) of X. We know that Mᵣ(X₂CO₃) = 2 × RAM(X) + RAM(C) + 3 × RAM(O). Using standard atomic masses: RAM(C) = 12.01 g/mol and RAM(O) = 16.00 g/mol. 138.42 = 2 × RAM(X) + 12.01 + (3 × 16.00) 138.42 = 2 × RAM(X) + 12.01 + 48.00 138.42 = 2 × RAM(X) + 60.01 2 × RAM(X) = 138.42 - 60.01 2 × RAM(X) = 78.41 RAM(X) = (78.41)/(2) RAM(X) = 39.21 g/mol What's next? Send 'em 📸