Question

The mass of $BaC{O}_3$ produced when excess of $C{O}_2$ is bubbled through a solution of $0.205\text{mole}Ba(OH{)}_2$ solution, is:
(Take molar mass of $Ba=137\text{g/mol}$)

$Ba(OH{)}_2+C{O}_2\to BaC{O}_3+H_{2}O$

• A. $81.63\text{g}$
• B. $20.25\text{g}$
• C. $40.38\text{g}$
• D. $55.95\text{g}$

Answer 1

The correct option is C $40.38\text{g}$

$Ba(OH{)}_2+C{O}_2\to BaC{O}_3+H_{2}O$

1 mol of $Ba(OH{)}_2$ gives 1 mol of $BaC{O}_3$.
So, 0.205 mol of $Ba(OH{)}_2$ will give 0.205 mol of $BaC{O}_3$.
mass of $BaC{O}_3$ formed = $\text{Moles}\times \text{Molar mass}$
Hence, mass of $BaC{O}_3$ formed $=0.205\text{mol}\times 197\text{g/mol}=40.38\text{g}$