The amount of BaSO$$_4$$ formed upon mixing 100 mL of 20.8% BaCl$$_2$$ solution with 50 mL of 9.8% H$$_2$$SO$$_4$$ solution will be: (Ba = 137, Cl = 35.5, S = 32, H = 1 and O = 16)

$$\text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{HCl}$$

$$\text{Weight of BaCl}_2 = 20.8\text{ g}; \text{ weight of H}_2\text{SO}_4 = 9.8\text{ }\% = 4.9\text{gm ,}$$

$$\begin{array}{ccccc} \text{BaCl}_2 & + & \text{H}_2\text{SO}_4 & \rightarrow & \text{BaSO}_4 & + & 2\text{HCl} \\ 0.1 & & 0.05 & & 0.05 & & \end{array}$$

$$\text{Weight of BaSO}_4 = 0.05 \times 233 = 11.65\text{ g}$$