Question

If 2 moles of $Ba(OH{)}_2$ reacts with 21 moles of $H_{3}P{O}_4$ then moles of product $B{a}_3(P{O}_4{)}_2$ formed are

• A. 21
• B. 7
• C. 6
• D. 14

Answer 1

The correct option is C 6

Given,

$2$ moles of $Ba(OH{)}_2$ reacts with $2L$ moles of $H_{3}P{O}_4$

The reaction involved is

$3Ba(OH{)}_2+2H_{3}P{O}_4⟶B{a}_3(P{O}_4{)}_2+6H_{2}O$

From above reaction, it is clear that

$3$ moles $Ba(OH{)}_2$ react with $2$ moles $H_{3}P{O}_4$ to give $1$ mole $B{a}_3(P{O}_4{)}_2$

$\Rightarrow 1$ moles $Ba(OH{)}_2$ consume $\frac{2}{3}$ moles $H_{3}P{O}_4$ to produce $\frac{1}{3}$ mole of $B{a}_3(P{O}_4{)}_2⟶(1)$

Here, $Ba(OH{)}_2$ is limiting reagent as only $2$ moles of $Ba(OH{)}_2$ is provided & $21$ moles of $H_{3}P{O}_4$

$\therefore$ From $\left(1\right)$

$1$ mole $Ba(OH{)}_2$ produce $\frac{1}{3}$ mole of $B{a}_3(P{O}_4{)}_2$

$\Rightarrow 2$ moles $Ba(OH{)}_2$ produce $\frac{2}{3}$ mole of $B{a}_3(P{O}_4{)}_2$

$\therefore$ Moles of $B{a}_3(P{O}_4{)}_2$ produced=$0.66$ moles

$\Rightarrow 0.66$ moles=$6\times {10}^{-1}$ moles

$=6$ deci moles