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Limiting Reagent or Reactant

If 0.5 moles ...

Question

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:
The corresponding balanced reaction is: $3 B a C l_{2} + 2 N a_{3} P O_{4} \to B a_{3} (P O_{4})_{2} + 6 N a C l$

A

0.7 mol

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B

0.5 mol

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C

0.2 mol

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D

0.1 mol

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Solution

The correct option is D 0.1 mol
The given balanced reaction is,
$3 B a C l_{2} + 2 N a_{3} P O_{4} \to B a_{3} (P O_{4})_{2} + 6 N a C l$
So, from the balanced reaction,
3 mol of $B a C l_{2}$ reacts with 2 mol of $N a_{3} P O_{4}$ to give 1 mol of $B a_{3} (P O_{4})_{2}$
The amount of $B a C l_{2}$ given here is 0.5 mol.
$∴$ The amount of $N a_{3} P O_{4}$ needed to completely react with 0.5 mol $B a C l_{2}$
$= \frac{2}{3} \times 0.5$ mol = 0.33 mol
But the amount of $N a_{3} P O_{4}$ given is 0.2 mol. Thus, $N a_{3} P O_{4}$ is the limiting reagent here.
Hence, the maximum number of moles of $B a_{3} (P O_{4})_{2}$ that can be formed:
$\frac{1}{2} \times 0.2$ mol = 0.1 mol

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Similar questions

Q. 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:
The corresponding balanced 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

0.3

B a C l_{2}

0.2

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_{2}

0.5

{B a C l}_{2}

0.2

{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

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B a C l_{2}

0.2

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that can be formed is:

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

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

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B a C l_{2}

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Limiting Reagent or Reactant

Standard XIII Chemistry

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