Question

If 0.5 moles of $BaC{l}_2$ is mixed with 0.2 moles 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:
The corresponding balanced reaction is: $3BaC{l}_2+2N{a}_{3}P{O}_4\to B{a}_3(P{O}_4{)}_2+6NaCl$

• 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 given balanced reaction is,
$3BaC{l}_2+2N{a}_{3}P{O}_4\to B{a}_3(P{O}_4{)}_2+6NaCl$
So, from the balanced reaction,
3 mol of $BaC{l}_2$ reacts with 2 mol of $N{a}_{3}P{O}_4$ to give 1 mol of $B{a}_3(P{O}_4{)}_2$
The amount of $BaC{l}_2$ given here is 0.5 mol.
$\therefore$The amount of $N{a}_{3}P{O}_4$ needed to completely react with 0.5 mol $BaC{l}_2$
$=\frac{2}{3}\times 0.5$ mol = 0.33 mol
But the amount of $N{a}_{3}P{O}_4$ given is 0.2 mol. Thus, $N{a}_{3}P{O}_4$ is the limiting reagent here.
Hence, the maximum number of moles of $B{a}_3(P{O}_4{)}_2$ that can be formed:
$\frac{1}{2}\times 0.2$ mol = 0.1 mol