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

The Correct Option is 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 $Ba_3(PO_4)_2$
$\because \, \, \, \, 2 moles \, of \, Na_3PO_4 \, gives \, 1 \, mole \, Ba_3(PO_4)_2$ .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$