Question

If $0.5$ mol of $BaCl_2$ is mixed with $0.2$ mol of $Na_3PO_4$ the maximum number of moles of $Ba_3(PO_4 )_2$ that can be formed is

• A. 0.7
• B. 0.5
• C. 0.1
• D. 0.2

Answer 1

Answer: (C)

The balanced chemical reaction is
$3BaCl_2 + 2Na_3PO_4 \rightarrow \, Ba_3(PO_4)_2 + 6NaCl $
In this reaction, $3$ moles of$ BaCl_2$ combines with $2$ moles of $Na_3PO_4$
Hence, $0.5$ mole of of $BaCl_2$ require
$ = \frac{2}{3} \times 0.5 = 0.33 $ mole of $Na_3PO_4$
Since, available $Na_3PO_4 (0.2$ mole) is less than required mole $(0.33)$,
it is the limiting reactant and would determine the amount of product $Ba_3(PO_4)_2$
$\because 2$ moles of $Na_3PO_4 $ gives $1 $ mole $Ba_3(PO_4)_2$
$\therefore \, \, 0.2$ mole of $Na_3PO_4$ would give $ \frac{1}{2} \times 0.2$
$ = 0.1 \,$ mole $\, Ba_3 (PO_4)_2$