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

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0.003 M of excess $C a^{2 +}$ will remain in excess

$C a {(N O_{3})}_{2} i s$ the excess reagent

0.003 moles of calcium oxalate will get precipitated

$N a_{2} C_{2} O_{4}$ is the limiting reagent

answer is A, C, D.

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Detailed Solution

$(a, c, d)$
$C a {(N O_{3})}_{2} + N a_{2} C_{2} O_{4} \to C a C_{2} O_{4} \to C a C_{2} O_{4} ↓ + 2 N a N O_{3}$
$100 \times 0.06 50 \times 0.06$
$= 6 m m o l = 3 m m o l 3 m m o l = 0. 003 m o l$
$N a_{2} C_{2} O_{4} i s t h e \lim i t i n g r e a g n e t$
$∴ 3 m m o l N a_{2} C_{2} O_{4} \equiv 3 m m o l C a {(N O_{3})}_{2} \equiv 3 m m o l C a C_{2} O_{4} \equiv 6 m m o l N a N O_{3}$
$m m o l o f C a {(N O_{3})}_{2} l e f t = 6 - 3 = m m o l = 0.003 m o l$
$M_{c a^{2 +}} (l e f t) = \frac{3 m m o l}{(100 + 50) m L} = \frac{3}{150} = 0.02 M$
$H e n c e, o p t i o n (b) i s w r o n g .$