For the reaction

$\mathrm{Ba}{\left(\mathrm{OH}\right)}_2+2{\mathrm{HClO}}_3\to \mathrm{Ba}{\left({\mathrm{ClO}}_3\right)}_2+2{\mathrm{H}}_2\mathrm{O}$, calculate the no. of moles of ${\mathrm{H}}_2\mathrm{O}$ formed when 0.1 mole of $\mathrm{Ba}{\left(\mathrm{OH}\right)}_2$ is treated with 0.0250 moles of ${\mathrm{HClO}}_3$

Given balanced chemical reaction:$\mathrm{Ba}{\left(\mathrm{OH}\right)}_2+2{\mathrm{HClO}}_3\to \mathrm{Ba}{\left({\mathrm{ClO}}_3\right)}_2+2{\mathrm{H}}_2\mathrm{O}$

$$\begin{gathered} 1 \mathrm{mole} \mathrm{of} \mathrm{Ba}{\left(\mathrm{OH}\right)}_2=2 \mathrm{mole} \mathrm{of} {\mathrm{H}}_2\mathrm{O} \\ 0.1 \mathrm{of} \mathrm{Ba}{\left(\mathrm{OH}\right)}_2 \mathrm{mole}=\frac{2 \mathrm{mole} \mathrm{of} {\mathrm{H}}_2\mathrm{O}}{1 \mathrm{mole} \mathrm{of} \mathrm{Ba}{\left(\mathrm{OH}\right)}_2}\times 0.1 \mathrm{mole} \mathrm{of} \mathrm{Ba}{\left(\mathrm{OH}\right)}_2 \\ =0.2 \mathrm{mole} \mathrm{of} {\mathrm{H}}_2\mathrm{O} \end{gathered}$$

$$\begin{gathered} 2 \mathrm{mole} \mathrm{of} {\mathrm{HClO}}_3=2 \mathrm{mole} {\mathrm{H}}_2\mathrm{O} \\ 0.025 \mathrm{mole} \mathrm{of} {\mathrm{HClO}}_3=\frac{2 \mathrm{mole} \mathrm{of} {\mathrm{H}}_2\mathrm{O}}{2 \mathrm{mole} \mathrm{of} {\mathrm{HClO}}_3}\times 0.025 \mathrm{mole} \mathrm{of} {\mathrm{HClO}}_3 \\ =0.025 \mathrm{mole} {\mathrm{H}}_2\mathrm{O} \end{gathered}$$

${\mathrm{HClO}}_3$ is limiting reagent, hence 0.025 mole of ${\mathrm{H}}_2\mathrm{O}$ will be formed.