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100 ml "of" 0.06 M Ca (NO(3))(2) is adde...

`100 ml "of" 0.06 M Ca (NO_(3))_(2)` is added to `50 mL` of `0.06 M Na_(2)C_(2) O_(4)`. After the reaction is complete.

A

0.003 moles of calcium oxalate will get precipitated

B

0.003 M `Ca^(2+)` will remain in excess

C

`Na_(2)C_(2)O_(4)` is the limiting reagent

D

Oxalate ion `(C_(2)O_(4)^(2))` concentration in final solution is 0.003 M

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The correct Answer is:
A, C
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