Question

If 0.59 mole of $BaC{l}_2$ is mixed with 0.20 mole 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. 0.7
• B. 0.5
• C. 0.2
• D. 0.1

Answer 1

The correct option is D 0.1

$3BaC{l}_2+2N{a}_{3}P{O}_4\to B{a}_3(P{O}_4{)}_2+6NaCl$

$3$ mole of $BaC{l}_2$ need $2$ mole of $N{a}_{3}P{O}_4$

$0.59$ mole of $BaC{l}_2$ needs $\frac{2}{3}\times 0.59N{a}_{3}P{O}_4$

$\Rightarrow$ as only $0.20$ mole of $N{a}_{3}P{O}_4$ are available decides the no. of moles of product as follows

$\Rightarrow 2N{a}_{3}P{O}_4$ gives $1$ mole of $B{a}_3(P{O}_4{)}_2$

$0.20$ mole gives $\frac{1}{2}\times 0.20$ moles of $B{a}_3(P{O}_4{)}_2$

$=0.10$ moles of $B{a}_3(P{O}_4{)}_2$