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.

Options

A. `0.003` moles of calcium oxalate will get precipitated.
B. `0.003M` of excess of `Ca^(2+)` will remain in excess
C. `Na_(2)C_(2)O_(4)` is limiting agent
D. `Ca(NO_(3))_(2)` is excess reagent

Answer

A, C, D

Explanation

`Ca^(2+) +C_(2)O_(4)^(2-) rarr CaC_(2)O_(4)`
10ml of `0.06M Ca(NO_(3))_(2)` contain `100 xx 0.06 = 6` millimoles
50ml of `0.06M Na_(2)C_(2)O_(4)` contains
`50 xx 0.06 = 3` millimoles
3 millimoles of `C_(2)O_(4)^(2-)` will react with 3 millmoles of `Ca^(2+)` for 3 millimoles of `CaC_(2)O_(4)`
`= (3)/(1000) = 0.003` moles (A) is correct Excess of `Ca^(2+)` left `= 6 - 3 = 3` millimoles `= (3)/(1000) = 0.003` moles.
Molarity of excess `Ca^(2+) = (3 xx 10^(-3))/(0.15) = 0.02M`
(C) is correct because sodium oxalate is present in smaller amount than required.
(D) is correct because `Ca(NO_(3))_(2)` is present in large amount.