The mass of BaCO₃ produced, when excess CO₂ is bubbled through a solution of 0.205 moles of Ba(OH)₂ is

$\mathrm{Ba}(\mathrm{OH}{)}_2+{\mathrm{CO}}_2\to {\mathrm{BaCO}}_3+{\mathrm{H}}_2\mathrm{O}$
$\text{ Atomic wt. of }{\mathrm{BaCO}}_3=137+12+16\times 3=197$
$\text{ No. of mole }=\frac{\text{ wt. of substance }}{\text{ mol wt. }}$
$\because 1\text{ mole of }\mathrm{Ba}(\mathrm{OH}{)}_2\text{ gives }1\text{ mole of }{\mathrm{BaCO}}_3$
$\therefore 0.205\text{ mole of }\mathrm{Ba}(\mathrm{OH}{)}_2\text{ will give }0.205{\text{ mole of BaCO}}_{\text{3}}$
$\therefore \text{ wt. of }0.205\text{ mole of }{\mathrm{BaCO}}_3\text{ will be }$
$0.205\times 197=40.385\mathrm{gm}\approx 40.5\mathrm{gm}$