Question

0.006 moles of $Ca(N{O}_3{)}_2$ are added to an aqueous solution containing 0.003 moles of $N{a}_2{C}_2{O}_4$ and the reaction is allowed to go to completion. Select the incorrect statement from the following:

• A. 0.003 mole of calcium oxalate will get precipitated.
• B. 0.001 moles of $C{a}^{2+}$ will remain in excess.
• C. $N{a}_2{C}_2{O}_4$ is the limiting reagent.
• D. $Ca(N{O}_3{)}_2$ is the excess reagent.

Answer 1

The correct option is B 0.001 moles of $C{a}^{2+}$ will remain in excess.

Moles of $Ca(N{O}_3{)}_2=0.006$ mole
Moles of $N{a}_2{C}_2{O}_4=0.003$ mole
$Ca(N{O}_3{)}_2+N{a}_2{C}_2{O}_4\to Ca(C_2{O}_4)\downarrow +2NaN{O}_3$
Limiting reagent will be $N{a}_2{C}_2{O}_4$ as the number of moles of $N{a}_2{C}_2{O}_4$ are lesser than required to completely react with $Ca(N{O}_3{)}_2.$
1 mole of $N{a}_2{C}_2{O}_4$ produces 1 mole of
of $Ca\left(C_2{O}_4\right)$
Hence, 0.003 moles of $N{a}_2{C}_2{O}_4$ will produce 0.003 moles of $Ca\left(C_2{O}_4\right)$
$Ca(N{O}_3{)}_2$ left = 0.006 - 0.003 = 0.003 mole