$100$ ml of $0.06$ M $C a (N O_{3})_{2}$ is added to $50$ ml of $0.06$ M $N a_{2} C_{2} O_{4}$ . After the reaction is complete :

0.003

moles of calcium oxalate will get precipitated.

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0.003

M of excess

C a^{2 +}

will remain in excess.

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

is the limiting reagent.

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

is the excess reagent.

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Solution

The correct options are
B $0.003$ moles of calcium oxalate will get precipitated.
C $C a (N O_{3})_{2}$ is the excess reagent.
D $N a_{2} C_{2} O_{4}$ is the limiting reagent.
$C a (N O_{3})_{2} 100 \times 0.06 = 6 m m o l + N a_{2} C_{2} O_{4} 50 \times 0.06 = 3 m m o l \to C a C_{2} O_{4} ↓ - 3 m m o l = 0.003 m o l + 2 N a N O_{3}$
$N a_{2} C_{2} O_{4}$ is the limiting reagent.
$∴ 3 m m o l N a_{2} C_{2} O_{4} = 3 m m o l C a (N O_{3})_{2}$
$\equiv 3 m m o l C a C_{2} O_{4} \equiv 6 m m o l N a N O_{3}$
mmol of $C a (N O_{3})_{2} l e f t = 6 - 3 = 3 m m o l = 0.003 m o l$
$M_{C a^{2 +}} (l e f t) = \frac{3 m m o l}{(100 + 50) m L} = \frac{3}{150} = 0.02 M$
Therefore, option B is wrong.
Hence, options A,C and D are correct.

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