The mass of ${\mathrm{BaCO}}_3$ produced when excess ${\mathrm{CO}}_2$ is bubbled through a solution of 0.205 mol $\mathrm{Ba}{\left(\mathrm{OH}\right)}_2$ is

$\begin{array}{l}\mathrm{Ba}{\left(\mathrm{OH}\right)}_2+{\mathrm{CO}}_2\to {\mathrm{BaCO}}_3+{\mathrm{H}}_2\mathrm{O}\\ \text{molecular\hspace{0.17em}}\mathrm{wt}.\mathrm{of}{\mathrm{BaCO}}_3=137+12+16\times 3=197\\ \mathrm{No}.\mathrm{of}\mathrm{mole}=\frac{\mathrm{wt}.\mathrm{of}\mathrm{substance}}{\mathrm{mol}\mathrm{wt}.}\\ \because 1\mathrm{mole}\mathrm{of}\mathrm{Ba}{\left(\mathrm{OH}\right)}_2\mathrm{gives}1\mathrm{mole}\mathrm{of}{\mathrm{BaCO}}_3\\ \therefore 0.205\mathrm{moles}\mathrm{of}\mathrm{Ba}{\left(\mathrm{OH}\right)}_2\mathrm{will}\mathrm{give}0.205\mathrm{mole}\mathrm{of}{\mathrm{BaCO}}_3\\ \therefore \mathrm{wt}.\mathrm{of}0.205\mathrm{mole}\mathrm{of}{\mathrm{BaCO}}_3\mathrm{will}\mathrm{be}\\ 0.205\times 197=40.385\mathrm{g}\approx 40.5\mathrm{g}\end{array}$