Explanation

(d)
$\begin{matrix}
3BaCl_{2} \\
3\ mol
\end{matrix} + \begin{matrix}
2Na_{3}PO_{4} \\
2\ mol
\end{matrix} \longrightarrow \begin{matrix}
Ba_{3}\left( PO_{4} \right)_{2} \\
1\ mol
\end{matrix} + 6NaCl$
Here, $Na_{3}PO_{4}$ is the limiting reactant.
2 moles of $Na_{3}PO_{4}$ gives 1 mole of $Ba_{3}\left( PO_{4} \right)_{2}$
So, 0.2 mole of $Na_{3}PO_{4}$ will give 0.1 mole of$Ba_{3}\left( PO_{4} \right)_{2}$.