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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 precipated

B

0.03 M of excess `Ca^(2+)` will remains in excess.

C

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

D

`Ca(NO_(3))_(2)` is the excess reagent.

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
`underset({:(100 xx 0.06,50xx0.06, -),(= 6 "mol",= 3 "mol", 3 "mmol = 0.003 "mol"):})(Ca(NO_(3))_(2) + Na_(2) C_(2) O_(4) rarr CaC_(2)O_(4)darr) + 2NANO_(3)`
`Na_(2) C_(2) O_(4)` is the limiting reagent.
`:.` 3 mmol `Na_(2) C_(2) O_(4) -= 2 "mmol" Ca(NO_(3))_(2)`
`-= 3 "mmol" CaC_(2) O_(4) -= 6 "mmol" NaNO_(3)`
mmol of `Ca(NO_(3))_(2)` left `= 6 - 3 = 3 "mmol" = 0.003 "mol"`
`M_(Ca^(2+))` (left) `= (3 "mmol")/((100 + 50) mL) = (3)/(150) = 0.02 M`
Hence, option (b) is wrong.
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