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Limiting Reagent or Reactant

0.006 moles o...

Question

0.006 moles of $C a (N O_{3})_{2}$ are added to an aqueous solution containing 0.003 moles of $N a_{2} C_{2} O_{4}$ and the reaction is allowed to go to completion. Select the incorrect statement from the following:

A

0.003 mole of calcium oxalate will get precipitated.

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B

0.001 moles of

C a^{2 +}

will remain in excess.

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C

N a_{2} C_{2} O_{4}

is the limiting reagent.

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D

C a (N O_{3})_{2}

is the excess reagent.

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Solution

The correct option is B 0.001 moles of $C a^{2 +}$ will remain in excess.
Moles of $C a (N O_{3})_{2} = 0.006$ mole
Moles of $N a_{2} C_{2} O_{4} = 0.003$ mole
$C a (N O_{3})_{2} + N a_{2} C_{2} O_{4} \to C a (C_{2} O_{4}) ↓ + 2 N a N O_{3}$
Limiting reagent will be $N a_{2} C_{2} O_{4}$ as the number of moles of $N a_{2} C_{2} O_{4}$ are lesser than required to completely react with $C a (N O_{3})_{2} .$
1 mole of $N a_{2} C_{2} O_{4}$ produces 1 mole of
of $C a (C_{2} O_{4})$
Hence, 0.003 moles of $N a_{2} C_{2} O_{4}$ will produce 0.003 moles of $C a (C_{2} O_{4})$
$C a (N O_{3})_{2}$ left = 0.006 - 0.003 = 0.003 mole

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Standard XII Chemistry

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