Question

If $0.5$ moles of $Ba{Cl}_2$ is mixed with $0.2$ moles of ${Na}_3{PO}_4$, the maximum amount of ${Ba}_3({PO}_4{)}_2$ that can be formed is :

• A. $0.7$ mol
• B. $0.5$ mol
• C. $0.2$ mol
• D. $0.1$ mol

Answer 1

The correct option is D $0.1$ mol

The balanced chemical reaction is $3BaC{l}_2+2N{a}_{3}P{O}_4\to B{a}_3(P{O}_4{)}_2+6NaCl$.
$0.5$ moles of barium chloride will require $\frac{0.5\times 2}{3}=0.33$ moles of sodium phosphate.
However, only $0.2$ moles of sodium phosphate are present.
Thus, sodium phosphate is the limiting reagent.
$2$ moles of sodium phosphate produces $1$ mole of barium phosphate.
Thus, $0.2$ moles of sodium phosphate will produce $0.1$ mole of barium phosphate.