Problem 148 - A Complete Resource Book in Chemistry for JEE Main

Question

\(100 \mathrm{ml}\) of \(0.06 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(50 \mathrm{~mL}\) of \(0.06\) \(\mathrm{M} \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\). After the reaction is complete (a) \(0.003 \mathrm{M}\) of excess of \(\mathrm{Ca}^{2+}\) will remain in excess. (b) \(0.003\) moles of calcium oxalate will get precipitated (c) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) is excess reagent. (d) \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) is limiting reagent.

Step-by-Step Solution

Answer

(b) and (d) are correct: 0.003 moles of calcium oxalate precipitate, and Na2C2O4 is the limiting reagent.

Step 1: Write the balanced chemical equation

The reaction between calcium nitrate, \( \mathrm{Ca(NO}_3)_2 \), and sodium oxalate, \( \mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4 \), produces calcium oxalate, \( \mathrm{CaC}_2 \mathrm{O}_4 \), and sodium nitrate, \( \mathrm{NaNO}_3 \). The balanced equation is: \[ \mathrm{Ca(NO}_3)_2 + \mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4 \rightarrow \mathrm{CaC}_2 \mathrm{O}_4 + 2\, \mathrm{NaNO}_3 \].\

Step 2: Calculate initial moles of each reactant

Calculate the moles of \( \mathrm{Ca(NO}_3)_2 \) and \( \mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4 \) using their molarities and volumes.- Moles of \( \mathrm{Ca(NO}_3)_2 \): \( 100 \, \mathrm{mL} \times 0.06 \, \mathrm{M} = 0.006 \, \text{moles} \).- Moles of \( \mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4 \): \( 50 \, \mathrm{mL} \times 0.06 \, \mathrm{M} = 0.003 \, \text{moles} \).

Step 3: Determine the limiting reagent

The stoichiometry of the reaction is 1:1. Since \( 0.006 \) moles of \( \mathrm{Ca(NO}_3)_2 \) react with 0.006 moles of \( \mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4 \) in stoichiometric amounts, but we only have 0.003 moles of \( \mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4 \), \( \mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4 \) is the limiting reagent.

Step 4: Calculate the moles of calcium oxalate formed

Since \( \mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4 \) is the limiting reagent with 0.003 moles, the reaction will form 0.003 moles of \( \mathrm{CaC}_2 \mathrm{O}_4 \) according to the balanced equation.

Step 5: Calculate the excess calcium ions

Initially, there were 0.006 moles of \( \mathrm{Ca(NO}_3)_2 \). Since only 0.003 moles react to form calcium oxalate, the excess moles of \( \mathrm{Ca}^{2+} \) are \( 0.006 - 0.003 = 0.003 \) moles.

Key Concepts

Limiting Reactant

In any chemical reaction, the concept of the limiting reactant is crucial. The limiting reactant is the substance that is completely used up first during the chemical reaction. It dictates when the reaction stops because no more products can be formed once the limiting reactant is gone.
To identify the limiting reactant, you need to compare the mole ratio of the reactants used in the reaction. This can be done once you have the balanced chemical equation. In our example, the reaction between calcium nitrate, \( \text{Ca(NO}_3)_2 \), and sodium oxalate, \( \text{Na}_2\text{C}_2\text{O}_4 \), is a 1:1 ratio.
Even though you start with 0.006 moles of \( \text{Ca(NO}_3)_2 \) and 0.003 moles of \( \text{Na}_2\text{C}_2\text{O}_4 \), since \( \text{Na}_2\text{C}_2\text{O}_4 \) is present in smaller amounts, it becomes the limiting reactant. This means that the reaction can only produce as much calcium oxalate as there is sodium oxalate available.
Identifying the limiting reactant is essential for accurately predicting the amount of product formed and determining any excess reactants left behind.

Chemical Reactions

Chemical reactions are processes where reactants convert to products. Chemical reactions involve rearrangement of atoms, and they can be represented by chemical equations.
Each chemical reaction entails breaking of chemical bonds in reactants and the formation of new bonds in products. This transformation follows the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction.
In the specific reaction between \( \text{Ca(NO}_3)_2 \) and \( \text{Na}_2\text{C}_2\text{O}_4 \), calcium oxalate \( \text{CaC}_2\text{O}_4 \) precipitates as a solid product. Precipitation reactions are a type of chemical reaction where an insoluble solid forms and separates from the solution. This particular reaction results in the formation of calcium oxalate that is insoluble in water, causing it to precipitate out of the solution.
Understanding chemical reactions allows chemists to predict product formation and also manipulate conditions to favor the desired products.

Balanced Chemical Equation

Balanced chemical equations are fundamental in stoichiometry, a branch of chemistry focused on the quantitative relationships between reactants and products in chemical reactions.
Balancing a chemical equation ensures that the same number of each type of atom appears on both sides of the equation, which adheres to the conservation of mass.
For the reaction between \( \text{Ca(NO}_3)_2 \) and \( \text{Na}_2\text{C}_2\text{O}_4 \), the balanced equation is:\[\text{Ca(NO}_3)_2 + \text{Na}_2\text{C}_2\text{O}_4 \rightarrow \text{CaC}_2\text{O}_4 + 2\, \text{NaNO}_3\]
This equation indicates that one mole of calcium nitrate reacts with one mole of sodium oxalate to form one mole of calcium oxalate and two moles of sodium nitrate.
Balancing chemical equations is a crucial step in any chemical calculation, as it provides the ratio of moles of reactants needed to form products, which further helps in identifying the limiting reactant, predicting the amounts of products formed, and understanding the stoichiometric relationships in the reaction.