Chapter 9, Problem 41A

Interpretation Introduction

Interpretation: The complete precipitation of calcium ion from the sample of 15 g of calcium ion needs to be determined, if 15 g of sodium oxalate is added.

Concept Introduction: A chemical reaction involves the conversion of reactant to product by breaking and making of chemical bonds.

There can be more than one reactant molecules involve in a chemical reaction to form the products. The reactant which is present in limited amount in a chemical reaction is called as limiting reactant.

Answer to Problem 41A

Since moles of sodium oxalate is lesser than moles of ${\text{Ca}}_{{}_{\text{(aq)}}}^{\text{2+}}$ ; thus 15 g of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4(aq)}}$ cannot precipitate all ${\text{Ca}}_{{}_{\text{(aq)}}}^{\text{2+}}$ ions from the solution.

Explanation of Solution

Given information:

  ${\text{Ca}}_{{}_{\text{(aq)}}}^{\text{2+}}{\text{+Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4(aq)}}\stackrel{}{\to }{\text{CaC}}_{\text{2}}{\text{O}}_{\text{4(aq)}}{\text{ + 2Na}}_{{}_{\text{(aq)}}}^{\text{2+}}$

Mass of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4(aq)}}$ = 15 g

Mass of ${\text{Ca}}_{{}_{\text{(aq)}}}^{\text{2+}}$ = 15 g

Molar mass of ${\text{Ca}}_{{}_{\text{(aq)}}}^{\text{2+}}$ = 40.0 g/mol

Molar mass of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4(aq)}}$ = 134 g/mol

Calculate moles of each:

  $\begin{array}{l}{\text{moles of Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4(aq)}}\text{ = }\frac{\text{15}\text{.00g}}{\text{134 g/mol}}\text{=0}\text{.112 moles}\\ {\text{moles of Ca}}_{{}_{}}^{2+}\text{ = }\frac{\text{15}\text{.00g}}{\text{40}\text{.0 g/mol}}\text{= 0}\text{.375 moles}\end{array}$

From the balance chemical equation:

1 mole of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4(aq)}}$ = 1 mole of ${\text{Ca}}_{{}_{\text{(aq)}}}^{\text{2+}}$

Since moles of sodium oxalate is lesser than moles of ${\text{Ca}}_{{}_{\text{(aq)}}}^{\text{2+}}$ ; thus 15 g of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4(aq)}}$ cannot precipitate all ${\text{Ca}}_{{}_{\text{(aq)}}}^{\text{2+}}$ ions from the solution.

Conclusion

Since moles of sodium oxalate is lesser than moles of ${\text{Ca}}_{{}_{\text{(aq)}}}^{\text{2+}}$ ; thus 15 g of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4(aq)}}$ cannot precipitate all ${\text{Ca}}_{{}_{\text{(aq)}}}^{\text{2+}}$ ions from the solution.