If $0.5$ moles of ${B a C l}_{2}$ 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 :
${3 B a C l}_{2} + {2 N a}_{3} {P O}_{4} ⟶ {B a}_{3} {(P O}_{4})_{2} + 6 N a C l$ .

0.7

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0.5

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0.3

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0.1

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Solution

The correct option is D $0.1$
The balanced chemical reaction is $3 B a C l_{2} + 2 N a_{3} P O_{4} \to B a_{3} (P O_{4})_{2} + 6 N a C l$ .
$0.5$ moles of barium chloride will require $\frac{0.5 \times 2}{3} = 0.33$ moles of sodium phosphate.
However, only $0.2$ moles of sodium phosphate are present.
Thus, sodium phosphate is the limiting reagent.
$2$ moles of sodium phosphate produce $1$ mole of barium phosphate.
Thus, $0.2$ moles of sodium phosphate will produce $0.1$ mole of barium phosphate.

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Similar questions

0.5

B a {C l}_{2}

0.2

{N a}_{3} {P O}_{4}

{B a}_{3} ({P O}_{4})_{2}

B a C l_{2}

N a_{3} P O_{4},

B a_{3} (P O_{4})_{2}

3 B a C l_{2} + 2 N a_{3} P O_{4} \to B a_{3} (P O_{4})_{2} + 6 N a C l

B a C l_{2}

N a_{3} P O_{4},

B a_{3} (P O_{4})_{2}

3 B a C l_{2} + 2 N a_{3} P O_{4} \to B a_{3} (P O_{4})_{2} + 6 N a C l

If 0.5 mole of BaCl₂ is mixed with 0.2 mole of Na₃PO₄,then maximum number of moles of Ba₃(PO₄)₂that can be formed

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