If 5 moles of BaCl2 is mixed with 2 moles of Na3PO4, the maximum number of moles of Ba3(PO4)2 formed is :

3BaCl2+2Na3PO4→Ba3PO42+6NaCl3 moles of BaCl2→2 moles of Na3PO4x→2 moles of Na3PO4⇒x=3 moles of BaCl2Limiting reagent is Na3PO42 moles of Na3PO4→1 mole of Ba3(PO4)2⇒1 mole of Ba3(PO4)2 will be formed

If 5 moles of BaCl2 is mixed with 2 moles of Na3PO4, the maximum number of moles of Ba3(PO4)2 formed is :

3BaCl2+2Na3PO4→Ba3PO42+6NaCl3 moles of BaCl2→2 moles of Na3PO4x→2 moles of Na3PO4⇒x=3 moles of BaCl2Limiting reagent is Na3PO42 moles of Na3PO4→1 mole of Ba3(PO4)2⇒1 mole of Ba3(PO4)2 will be formed

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If 5 moles of BaCl₂ is mixed with 2 moles of Na₃PO₄, the maximum number of moles of Ba₃(PO₄)₂ formed is :

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1.8

3.23

4.25

answer is C.

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Detailed Solution

$\begin{array}{l} 3 {BaCl}_{2} + 2 {Na}_{3} {PO}_{4} \to {Ba}_{3} {({PO}_{4})}_{2} + 6 NaCl \\ 3 moles of {BaCl}_{2} \to 2 moles of {Na}_{3} {PO}_{4} \\ x \to 2 moles of {Na}_{3} {PO}_{4} \\ \Rightarrow x = 3 moles of {BaCl}_{2} \end{array}$
Limiting reagent is ${Na}_{3} {PO}_{4}$
2 moles of ${Na}_{3} {PO}_{4} \to$ 1 mole of ${Ba}_{3} {({PO}_{4})}_{2}$
$\Rightarrow$ 1 mole of ${Ba}_{3} {({PO}_{4})}_{2}$ will be formed