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

Amount of ${\mathrm{BaCl}}_2$in $100 \mathrm{mL}$ of $20.8\% {\mathrm{BaCl}}_2$ $=\frac{m}{M}=\frac{20.8\mathrm{g}}{(137+2\times 35.5)\mathrm{g}{\mathrm{mol}}^{-1}}=0.1\mathrm{mol}$

Amount of${\mathrm{H}}_2{\mathrm{SO}}_4$ in $50 \mathrm{mL} \mathrm{of} 9.8\% {\mathrm{H}}_2{\mathrm{SO}}_4$

$\equiv \frac{m}{M}=\frac{(9.8/2)\mathrm{g}}{98\mathrm{g}{\mathrm{mol}}^{-1}}=0.05\mathrm{mol}$

From the reaction $\underset{0.1 mol }{{\mathrm{BaCl}}_2}+\underset{0.05 \mathrm{mol}}{{\mathrm{H}}_2{\mathrm{SO}}_4}\to {\mathrm{BaSO}}_4+2\mathrm{HCl}$

We conclude that 0.05 mol of ${\mathrm{BaSO}}_4$ will be formed

Hence, $m\left({\mathrm{BaSO}}_4\right)=\left(0.05\mathrm{mol}\right)\left(233\mathrm{g}{\mathrm{mol}}^{-1}\right)=11.65\mathrm{g}$