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Limiting Reagent or Reactant

0.5 mole Ba...

Question

0.5 mole $B a C l_{2}$ is mixed with 0.2 mole $N a_{3} P O_{4}$ , the maximum number of moles of $B a_{3} (P O_{4})_{2}$ that can be formed is :

A

0.7

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B

0.5

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C

0.2

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D

0.1

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Solution

The correct option is D 0.1
Balance equation,

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

3 moles of $B a C l_{2}$ reacts with 2 moles of $N a_{3} P O_{4}$ .

Then 0.5 moles of $B a C l_{2}$ reacts with $2 \times \frac{0.5}{3} = 0.33$ moles of $N a_{3} P O_{4}$ .

But the amount of $N a_{3} P O_{4}$ present is less than 0.2 moles.

Hence, $N a_{3} P O_{4}$ is the limiting reagent and will decide the amount of product formed.
0.2 moles of $N a_{3} P O_{4}$ reacts with 0.3 moles of $B a C l_{2}$ and produces 0.1 mole of $B a_{3} (P O_{4})_{2}$

Hence, the correct option is $D$

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Q. If 0.5 moles of

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is mixed with 0.2 moles of

N a_{3} P O_{4},

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

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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}

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