Problem 41 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

What weight of \(\mathrm{AgCl}\) will be precipitated when a solution containing \(4.77 \mathrm{~g}\) of \(\mathrm{NaCl}\) is added to a solution of \(5.77 \mathrm{~g}\) of \(\mathrm{AgNO}_{3} ?\) [1978]

Step-by-Step Solution

Answer

4.87 g of AgCl will be precipitated.

Step 1: Write the Balanced Chemical Equation

The reaction between sodium chloride (\(\mathrm{NaCl}\)) and silver nitrate (\(\mathrm{AgNO_3}\)) produces silver chloride (\(\mathrm{AgCl}\)) and sodium nitrate (\(\mathrm{NaNO_3}\)) by the equation: \[\mathrm{NaCl} + \mathrm{AgNO_3} \rightarrow \mathrm{AgCl} + \mathrm{NaNO_3}.\]

Step 2: Determine the Molar Masses

Calculate the molar masses of the reactants and the product. - Molar mass of \(\mathrm{NaCl}\) is \(58.44 \text{ g/mol}\).- Molar mass of \(\mathrm{AgNO_3}\) is \(169.87 \text{ g/mol}\).- Molar mass of \(\mathrm{AgCl}\) is \(143.32 \text{ g/mol}\).

Step 3: Calculate Moles of NaCl and AgNO3

Use the formula \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\) to find the moles of \(\mathrm{NaCl}\) and \(\mathrm{AgNO_3}\).For \(\mathrm{NaCl}\):\[\text{Moles of NaCl} = \frac{4.77 \text{ g}}{58.44 \text{ g/mol}} \approx 0.0816 \text{ moles}.\]For \(\mathrm{AgNO_3}\):\[\text{Moles of AgNO_3} = \frac{5.77 \text{ g}}{169.87 \text{ g/mol}} \approx 0.0340 \text{ moles}.\]

Step 4: Identify the Limiting Reagent

In a 1:1 mole ratio reaction, the limiting reagent is the one present in fewer moles. Here, 0.0340 moles of \(\mathrm{AgNO_3}\) is lesser compared to 0.0816 moles of \(\mathrm{NaCl}\). Thus, \(\mathrm{AgNO_3}\) is the limiting reagent.

Step 5: Calculate the Amount of AgCl Formed

Since \(\mathrm{AgNO_3}\) is the limiting reagent, the amount of \(\mathrm{AgCl}\) precipitated is set by its moles. Hence, 0.0340 moles of \(\mathrm{AgCl}\) will form because of the 1:1 reaction ratio.Now, convert moles of \(\mathrm{AgCl}\) to grams:\[\text{Mass of AgCl} = 0.0340 \text{ moles} \times 143.32 \text{ g/mol} \approx 4.87 \text{ g}.\]

Key Concepts

Limiting Reagent

In any chemical reaction, the limiting reagent plays a pivotal role in determining the amount of product that can be formed. The limiting reagent is the reactant that is completely consumed first, halting the reaction because there are no more reactants left to continue the process.
In our example, we have a reaction between sodium chloride (\(\mathrm{NaCl}\)) and silver nitrate (\(\mathrm{AgNO_3}\)), which yields silver chloride (\(\mathrm{AgCl}\)) and sodium nitrate (\(\mathrm{NaNO_3}\)). To identify the limiting reagent, we compare the number of moles present of each reactant.

• The balanced equation indicates a 1:1 molar ratio between \(\mathrm{NaCl}\) and \(\mathrm{AgNO_3}\).
• Upon calculation, \(\mathrm{NaCl}\) provides 0.0816 moles and \(\mathrm{AgNO_3}\) offers 0.0340 moles.

Since \(\mathrm{AgNO_3}\) has fewer moles, it becomes the limiting reagent. Understanding this concept ensures predictions about the maximum amount of product, here \(\mathrm{AgCl}\), that the reaction can yield.

Molar Mass Calculation

Molar mass is the bridge that connects the mass of a substance to the number of moles, and thus, its molecules. For a compound, the molar mass is calculated by summing the atomic masses of all the atoms in the molecule. It's expressed in grams per mole (g/mol).
Let's break down the process for the substances involved in our exercise:

• \(\mathrm{NaCl}\): Sodium has a molar mass of about 22.99 g/mol, and chlorine weighs about 35.45 g/mol. Their combined weight gives \(\mathrm{NaCl}\) a molar mass of 58.44 g/mol.
• \(\mathrm{AgNO_3}\): Silver is 107.87 g/mol, nitrogen adds another 14.01 g/mol, and the three oxygen atoms combine to 48.00 g/mol, totaling \(\mathrm{AgNO_3}\) at 169.87 g/mol.
• \(\mathrm{AgCl}\): With silver and chlorine, \(\mathrm{AgCl}\)'s molar mass is 143.32 g/mol.

Understanding these calculations enables students to convert mass into moles, crucial for solving stoichiometry problems.

Chemical Equations

A chemical equation is a symbolic representation of a chemical reaction where reactants are on the left, and products are on the right. The equation must be balanced, showing the conservation of mass. This means the number of each type of atom on the reactants side equals that on the products side.
In the given problem, the balanced equation is:
\[ \mathrm{NaCl} + \mathrm{AgNO_3} \rightarrow \mathrm{AgCl} + \mathrm{NaNO_3}\]

• Each reactant and product has the same number of respective atoms: 1 sodium, 1 chlorine, 1 silver, 1 nitrogen, and 3 oxygens.
• The arrow indicates the direction of the reaction.

Balancing chemical equations ensures that mass is conserved according to the law of conservation of mass. It's fundamental for accurately determining relationships between quantities of reactants and products, a critical component in stoichiometry.