Chemistry - Mole Concept Question with Solution | TestHub

$\begin{array}{c}\text{3}{\text{BaCl}}_{\text{2}}\\ \text{3 mol}\end{array}\text{+}\begin{array}{c}\text{2}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\\ \text{2 mol}\end{array}\to \begin{array}{c}{\text{Ba}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\\ \text{1 mol}\end{array}\text{+}\text{6}\text{NaCl}$
Here,${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$is the limiting reactant.
$\text{2}$moles of${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$gives$\text{1}$mole of${\text{Ba}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$
So,$0.2$mole of${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$will give$\text{0}\text{.1}$mole of${\text{Ba}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$