This problem combines solution concentration, mole calculations, limiting reagent identification, and stoichiometry — a multi-concept JEE Advanced challenge. Step 1: calculate moles of each reactant. Moles of (\text{BaCl}_2 = 0.5 , \text{mol/L} \times 0.100 , \text{L} = 0.05) mol. Moles of (\text{Na}_2\text{SO}_4 = 0.3 , \text{mol/L} \times 0.200 , \text{L} = 0.06) mol. Step 2: identify the limiting reagent. The reaction is 1:1, so the reagent with fewer moles is limiting. (\text{BaCl}_2) (0.05 mol) < (\text{Na}_2\text{SO}_4) (0.06 mol), so (\text{BaCl}_2) is the limiting reagent. Step 3: calculate mass of (\text{BaSO}_4). From stoichiometry, 1 mol (\text{BaCl}_2) produces 1 mol (\text{BaSO}_4). Moles of (\text{BaSO}_4 = 0.05) mol. Mass (= 0.05 \times 233 = 11.65) g. Option 5.825 g corresponds to 0.025 mol, which is half the amount from 0.05 mol (\text{BaCl}_2) — a factor-of-2 error. Option 13.98 g would require 0.06 mol (\text{BaSO}_4), arising from incorrectly using (\text{Na}_2\text{SO}_4) (the excess reagent) as the limiting reagent. Option 23.3 g corresponds to 0.1 mol, which would be the theoretical maximum if 0.1 mol of either reagent were available — twice the actual (\text{BaCl}_2) moles. Plausibility check: 11.65 g of a dense barium salt precipitate from millimolar-scale solutions is physically reasonable, and 0.05 mol × 233 g/mol = 11.65 g with exact arithmetic.