A) 0.1
B) 0.2
C) 0.5
D) 0.7
Correct Answer: A
Solution :
\[\underset{[3\,\,\,mol}{\mathop{3BaC{{l}_{2}}}}\,+\underset{2\,\,\,mol}{\mathop{2N{{a}_{3}}P{{O}_{4}}}}\,\to \underset{6\,\,\,mol}{\mathop{6NaCl}}\,+\underset{1\,\,\,mol]}{\mathop{B{{a}_{3}}{{(P{{O}_{4}})}_{2}}}}\,\] Given \[\Rightarrow \]0.5 mol of \[BaC{{l}_{2}}\]and 0.2 mol of \[N{{a}_{3}}P{{O}_{4}}\] To find the limiting reagent: 2 mol of \[N{{a}_{3}}P{{O}_{4}}\]\[\Rightarrow \]3 mol of \[BaC{{l}_{2}}\] 0.2 mol of \[N{{a}_{3}}P{{O}_{4}}\] \[\Rightarrow \]0.3 mol of \[BaC{{l}_{2}}\] \[\therefore \] \[N{{a}_{3}}P{{O}_{4}}\]is the limiting reagent \[\therefore \] 2 mol of \[N{{a}_{3}}P{{O}_{4}}\]\[\Rightarrow \]1 mol of \[B{{a}_{3}}{{(P{{O}_{4}})}_{2}}\] 0.2 mol of \[N{{a}_{3}}P{{O}_{4}}\]\[\Rightarrow \]0.1 mol of \[B{{a}_{3}}{{(P{{O}_{4}})}_{2}}\]
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