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

Statements (c) and (d) are correct. Calcium nitrate is in excess; sodium oxalate is the limiting reagent.

Step 1: Identify the reaction

The reaction between calcium nitrate and sodium oxalate is:\[\text{Ca(NO}_3\text{)}_2 + \text{Na}_2\text{C}_2\text{O}_4 \rightarrow \text{CaC}_2\text{O}_4 \downarrow + 2\text{NaNO}_3\]This is a precipitation reaction where calcium oxalate precipitates out.

Step 2: Calculate the moles of reactants

Calculate the moles of each reactant:- For calcium nitrate, \[ \text{Moles of } \text{Ca(NO}_3\text{)}_2 = 0.06 \text{ M} \times 0.1 \text{ L} = 0.006 \text{ moles} \]- For sodium oxalate, \[ \text{Moles of } \text{Na}_2\text{C}_2\text{O}_4 = 0.06 \text{ M} \times 0.05 \text{ L} = 0.003 \text{ moles} \]

Step 3: Determine the limiting reagent

Both reactants react in a 1:1 mole ratio. Since there are 0.006 moles of calcium nitrate and only 0.003 moles of sodium oxalate, sodium oxalate is the limiting reagent.

Step 4: Calculate moles of precipitate formed

Since sodium oxalate is the limiting reagent with 0.003 moles, it will completely react, leading to the formation of 0.003 moles of calcium oxalate precipitate.

Step 5: Calculate remaining excess reagent

The limiting reagent (sodium oxalate) fully reacts, thus:- Remaining moles of calcium nitrate after reaction: \[ 0.006 - 0.003 = 0.003 \text{ moles} \]The concentration of excess calcium ions in 150 mL (0.15 L) of solution is:\[\text{Concentration} = \frac{0.003 \text{ moles}}{0.15 \text{ L}} = 0.02 \text{ M}\]

Key Concepts

Limiting Reagent

In a chemical reaction, the liming reagent is the substance that runs out first and thus limits the amount of products that can be formed. In our example, the reaction is between calcium nitrate

• Ca(NO₃)₂
• sodium oxalate, Na₂C₂O₄

To identify the limiting reagent, we compare the amount of each reactant using mole calculations and the stoichiometry of the balanced equation:\[ \text{Ca(NO}_3\text{)}_2 + \text{Na}_2\text{C}_2\text{O}_4 \rightarrow \text{CaC}_2\text{O}_4 \downarrow + 2\text{NaNO}_3 \]Because they react in a 1:1 ratio, we calculate which substance we have less of in terms of moles. In this reaction, you begin with

• 0.006 moles of calcium nitrate
• 0.003 moles of sodium oxalate

Although the calcium nitrate seems abundant, the sodium oxalate acts as the limiting reagent, allowing only 0.003 moles of product formation. This restraint halts other reactions once the sodium oxalate is depleted.

Precipitation Reaction

A precipitation reaction occurs when two aqueous solutions combine to form an insoluble product known as a precipitate. In the reaction we explored, calcium nitrate reacts with sodium oxalate to produce calcium oxalate:\[ \text{Ca(NO}_3\text{)}_2 + \text{Na}_2\text{C}_2\text{O}_4 \rightarrow \text{CaC}_2\text{O}_4 \downarrow + 2\text{NaNO}_3 \]The precipitate, calcium oxalate \(\text{CaC}_2\text{O}_4\), is solid and forms as a result of the low solubility of oxalate with calcium ions compared to sodium ions. Here's what happens:

• Calcium ions \(\text{Ca}^{2+}\) from calcium nitrate meet oxalate ions \(\text{C}_2\text{O}_4^{2-}\) from sodium oxalate in the solution.
• Their combination forms the insoluble calcium oxalate, which precipitates out of the solution.

This type of reaction is crucial for forming various compounds, especially when purifying or extracting materials. Precipitation reactions can be made to occur by slowly combining reactant solutions to minimize the formation of impurities.

Mole Calculations

Mole calculations are fundamental in determining how much of each reactant and product is involved in a chemical reaction. They help us understand the stoichiometry or the quantitative relationships in the reaction. In the exercise, we determine the moles of each substance based on the molarity and the volume:- \( \text{Moles of Ca(NO}_3\text{)}_2 = 0.06 \text{ M} \times 0.1 \text{ L} = 0.006 \text{ moles} \)- \( \text{Moles of Na}_2\text{C}_2\text{O}_4 = 0.06 \text{ M} \times 0.05 \text{ L} = 0.003 \text{ moles} \)By employing these calculations:

• It becomes easy to see that sodium oxalate is the limiting reagent.
• We can calculate the amount of each product.
• The remaining excess of calcium nitrate after sodium oxalate is fully consumed.

Understanding how to perform these calculations provides a clear picture of how reactions proceed and are essential for any thorough chemical analysis. They ensure that we are aware of the practical limits of chemical reactions and can calculate precisely how much product will be formed and what amounts of reactants will be left unreacted.