Question

100 mL of 0.06 M $Ca(N{O}_3{)}_2$ is added to 50 mL of 0.06 M $N{a}_2{C}_2{O}_4$. Which of the following statement(s) is/are correct?

• A. 0.003 mol of calcium oxalate will precipitate
• B. 0.003 M of excess of $C{a}^{2+}$ will remain in excess
• C. $N{a}_2{C}_2{O}_4$ is the limiting reagent
• D. $Ca(NO{)}_3$ is the excess reagent

Answer 1

Answer: (A), (C), (D)

The correct options are
A 0.003 mol of calcium oxalate will precipitate
C $N{a}_2{C}_2{O}_4$ is the limiting reagent
D $Ca(NO{)}_3$ is the excess reagent

Balanced reaction is : $Ca(N{O}_3{)}_2+N{a}_2{C}_2{O}_4\to 2NaN{O}_3+Ca{C}_2{O}_4$

We know, Molarity = $\frac{\text{number of moles}}{\text{volume of the solution in mL}}\times 1000$
For $Ca(N{O}_3{)}_2$, number of moles = $\frac{0.06\times 100}{1000}$ = $6\times {10}^{-3}$ mol
Similarly for $N{a}_2{C}_2{O}_4$, number of moles = $\frac{0.06\times 50}{1000}$ = $3\times {10}^{-3}$ mol
According to the reaction, 1 mol of calcium nitrate will react with 1 mol of sodium oxalate to form 1 mol of calcium oxalate.
$\therefore$ amount of calcium oxalate formed = $3\times {10}^{-3}$ mol
Calcium nitrate is the excess reagent.
moles of calcium nitrate left = $3\times {10}^{-3-+}$ mol
In calcium nitrate, dissociation will give 1 ion of $C{a}^{2+}$ and 2 ions of $N{O}_3^{-}$
Total volume = 150 mL
Molarity = $\frac{3\times {10}^{-3}}{150}\times 1000$ = 0.02 M of $C{a}^{2+}$