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Standard XI

Chemistry

Limiting Reagent or Reactant

If 0.5 mole o...

Question

If $0.5 m o l e$ of $B a C l_{2}$ is mixed with $0.2 m o l e$ of $N a_{3} P O_{4}$ , the maximum number of moles of $B a_{3} (P O_{4})_{2}$ formed are

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

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

Limiting reagent is $N a_{3} P O_{4}$

$2 m o l e N a_{3} P O_{4} \to 1 m o l e B a_{3} (P O_{4})_{2}$

$0.2 m o l e N a_{3} P O_{4} \to 0.1 m o l e B a_{3} (P O_{4})_{2}$

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Similar questions

Q. 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 :

If $0.5 m o l e$ of $B a C l_{2}$ is mixed with $0.2 m o l e$ of $N a_{3} P O_{4}$ , the maximum number of moles of $B a_{3} (P O_{4})_{2}$ formed are

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 moles of

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

obtained is

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

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

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If 0.5 mole of BaCl2 is mixed with 0.2 mole of Na3PO4, the maximum number of moles of Ba3(PO4)2 formed are

The correct option is D 0.1 3BaCl2 + 2Na3PO4 → Ba3(PO4)2 + 6NaCl Limiting reagent is Na3PO4 2 mole Na3PO4 → 1 mole Ba3(PO4)2 0.2 mole Na3PO4 → 0.1 mole Ba3(PO4)2

If $0.5 m o l e$ of $B a C l_{2}$ is mixed with $0.2 m o l e$ of $N a_{3} P O_{4}$ , the maximum number of moles of $B a_{3} (P O_{4})_{2}$ formed are

The correct option is D
0.1

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

Limiting reagent is $N a_{3} P O_{4}$

$2 m o l e N a_{3} P O_{4} \to 1 m o l e B a_{3} (P O_{4})_{2}$

$0.2 m o l e N a_{3} P O_{4} \to 0.1 m o l e B a_{3} (P O_{4})_{2}$