# If $0.50$ mol of $Bacl_2$ is mixed with $0.20$ mol of $Na_3PO_4$ , the maximum number of moles of $Ba_3(PO_4)_2$ that can be formed is

$(a)\;0.70 \\ (b)\;0.50 \\(c)\;0.20 \\(d)\;0.10$

The reaction can be represented as as $3 BaCl_2+2Na_3PO_4 \longrightarrow Ba_3(PO_4)+6NaCl$
From the above balanced equation, 3 moles of $Bacl_2$ required 2 moles of $Na_3PO_4$.
$0.5$ moles of $Bacl_2$ will require moles of $Na_3PO_4=\large\frac{2}{3} $$\times 0.5 =0.33 Since only 0.2 moles of Na_3PO_4 are available . So, Na_3PO_4 is the limiting reagent. Since 1 mole of Ba_3(PO_4) is formed when 2 moles of Na_3PO_4 react. So the moles of Ba_3(PO_4)_2 formed when .2 moles of Na_3PO_4 reacts =\large\frac{1}{2} \times$$ 0.2 =0.1$
Hence d is the correct answer.