Question
If 0.50 mole of $$BAC{l_{\text{2}}}$$ is mixed with 0.20 mol of $$N{a_3}P{O_4},$$ the maximum number of moles of $$B{A_3}\left( {P{O_4}} \right)$$ that can be formed is
A.
0.70
B.
0.50
C.
0.20
D.
0.10
Answer :
0.10
Solution :
TIPS/Formulae:
(i) Write balanced chemical equation for chemical change.
(ii) Find limiting reagent.
(iii) Amount of product formed will be determined by amount of limiting reagent.
The balanced equation is :
\[\underset{\begin{smallmatrix}
No.\,of \\
moles:
\end{smallmatrix}}{\mathop {}}\,\,\,\underset{\begin{smallmatrix}
\,\,\,0.5 \\
3\,mole
\end{smallmatrix}}{\mathop{3BaC{{l}_{2}}}}\,+\underset{\begin{smallmatrix}
\,\,\,0.2 \\
2\,mole
\end{smallmatrix}}{\mathop{2N{{a}_{3}}P{{O}_{4}}}}\,\to \underset{\begin{smallmatrix}
\\
1\,mole
\end{smallmatrix}}{\mathop{B{{a}_{3}}\left( P{{O}_{4}} \right)}}\,+6NaCl\]
Limiting reagent is $$N{a_3}P{O_4}\left( {0.2\,mol} \right),\,BaC{l_2}$$ is in excess.
From the above equation :
$$\eqalign{
& 2.0\,{\text{moles}}\,\,{\text{of}}\,N{a_3}P{O_4}\,{\text{yields}}\,\,B{a_3}{\left( {P{O_4}} \right)_2} = 1\,{\text{mole}} \cr
& \therefore {\text{0}}{\text{.2}}\,{\text{moles of}}\,N{a_3}P{O_4}\,{\text{will}}\,{\text{yields}}\,B{a_3}{\left( {P{O_4}} \right)_2} = \frac{1}{2} \times 0.2 \cr
& = 0.1\,{\text{mol}}{\text{.}} \cr} $$