Question

If $0.5$ mol of $BaC{l}_2$ is mixed with $0.2$ mol of $N{a}_{3}P{O}_4$ the maximum number of moles of $B{a}_3(P{O}_4{)}_2$ that can be formed is

• A. 0.7
• B. 0.5
• C. 0.1
• D. 0.2

Answer 1

Answer: (C)

The balanced chemical reaction is
$3BaC{l}_2+2N{a}_{3}P{O}_4\to B{a}_3(P{O}_4{)}_2+6NaCl$
In this reaction, $3$ moles of$BaC{l}_2$ combines with $2$ moles of $N{a}_{3}P{O}_4$
Hence, $0.5$ mole of of $BaC{l}_2$ require
$=\frac{2}{3}\times 0.5=0.33$ mole of $N{a}_{3}P{O}_4$
Since, available $N{a}_{3}P{O}_4(0.2$ mole) is less than required mole $(0.33)$,
it is the limiting reactant and would determine the amount of product $B{a}_3(P{O}_4{)}_2$
$\because 2$ moles of $N{a}_{3}P{O}_4$ gives $1$ mole $B{a}_3(P{O}_4{)}_2$
$\therefore 0.2$ mole of $N{a}_{3}P{O}_4$ would give $\frac{1}{2}\times 0.2$
$=0.1$ mole $B{a}_3(P{O}_4{)}_2$