Some Basic Concepts of Chemistry - Result Question 13
27. If 0.50 mole of $\mathrm{BaCl}_2$ is mixed with 0.20 mole of $\mathrm{Na}_3 \mathrm{PO}_4$, the maximum number of moles of $\mathrm{Ba}_3\left(\mathrm{PO}_4\right)_2$ that can be formed is
(1981, 1M)
(a) 0.70
(b) 0.50
(c) 0.20
(d) 0.10
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Answer:
Correct Answer: 27. (d)
Solution:
- The balanced chemical reaction is
$3 \mathrm{BaCl}_2+2 \mathrm{Na}_3 \mathrm{PO}_4 \longrightarrow \mathrm{Ba}_3\left(\mathrm{PO}_4\right)_2+6 \mathrm{NaCl}$
In this reaction, 3 moles of $\mathrm{BaCl}_2$ combines with 2 moles of $\mathrm{Na}_3 \mathrm{PO}_4$. Hence, 0.5 mole of of $\mathrm{BaCl}_2$ require
$\frac{2}{3} \times 0.5=0.33 \text { mole of } \mathrm{Na}_3 \mathrm{PO}_4 \text {. }$
Since, available $\mathrm{Na}_3 \mathrm{PO}_4(0.2 \mathrm{~mole})$ is less than required mole $(0.33)$, it is the limiting reactant and would determine the amount of product $\mathrm{Ba}_3\left(\mathrm{PO}_4\right)_2$.
$\because 2$ moles of $\mathrm{Na}_3 \mathrm{PO}_4$ gives 1 mole $\mathrm{Ba}_3\left(\mathrm{PO}_4\right)_2$
$\therefore 0.2$ mole of $\mathrm{Na}_3 \mathrm{PO}_4$ would give $\frac{1}{2} \times 0.2=0.1 \mathrm{~mole}^2 \mathrm{Ba}_3\left(\mathrm{PO}_4\right)_2$
BETA
