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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 $
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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.
answered Mar 21, 2014 by meena.p

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