Problem 16 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

If \(0.50\) mole of \(\mathrm{BaCl}_{2}\) is mixed with \(0.20 \mathrm{~mol}\) of \(\mathrm{Na}_{3} \mathrm{PO}_{4}\), the maximum number of moles of \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) that can be formed is [1981 - 1 Mark] (a) \(0.70\) (b) \(0.50\) (c) \(0.20\) (d) \(0.10\)

Step-by-Step Solution

Answer

(d) 0.10

Step 1: Write the balanced chemical equation

Identify the reaction between barium chloride \((\text{BaCl}_2)\) and sodium phosphate \((\text{Na}_3\text{PO}_4)\) to form barium phosphate \((\text{Ba}_3(\text{PO}_4)_2)\) and sodium chloride \((\text{NaCl})\). The balanced equation is:\[3\, \text{BaCl}_2 (aq) + 2\, \text{Na}_3\text{PO}_4 (aq) \rightarrow \text{Ba}_3(\text{PO}_4)_2 (s) + 6\, \text{NaCl} (aq)\].

Step 2: Determine mole ratio from the balanced equation

From the balanced equation, the mole ratio of \(\text{BaCl}_2\) to \(\text{Ba}_3(\text{PO}_4)_2\) is 3:1, and the mole ratio of \(\text{Na}_3\text{PO}_4\) to \(\text{Ba}_3(\text{PO}_4)_2\) is 2:1.

Step 3: Identify the limiting reactant

Calculating based on \(\text{BaCl}_2\) first: \(0.50\) moles of \(\text{BaCl}_2\) can potentially produce \(\frac{0.50}{3} = 0.1667\) moles of \(\text{Ba}_3(\text{PO}_4)_2\).Calculating based on \(\text{Na}_3\text{PO}_4\): \(0.20\) moles of \(\text{Na}_3\text{PO}_4\) can potentially produce \(\frac{0.20}{2} = 0.10\) moles of \(\text{Ba}_3(\text{PO}_4)_2\).The smaller amount of \(\text{Ba}_3(\text{PO}_4)_2\) calculated identifies \(\text{Na}_3\text{PO}_4\) as the limiting reactant.

Step 4: Calculate maximum moles of \(\text{Ba}_3(\text{PO}_4)_2\) formed

Since \(\text{Na}_3\text{PO}_4\) is the limiting reactant, the maximum number of moles of \(\text{Ba}_3(\text{PO}_4)_2\) that can be formed is \(0.10\) moles.

Key Concepts

Limiting Reactant

In chemical reactions, the concept of a limiting reactant is crucial for knowing how much product can be formed. When two or more reactants are mixed together, there's usually one reactant that gets used up before the others. This reactant is known as the limiting reactant. Once it's consumed completely, no more product can form because you need all reactants present to continue the reaction. Understanding which reactant is limiting is essential in predicting the maximum amount of product possible from a reaction.
For example, in our exercise, we mixed \( 0.50 \text{ mol } \ ext{BaCl}_2 \) with \( 0.20 \text{ mol } \ ext{Na}_3\text{PO}_4 \). We found that \( \ ext{Na}_3\text{PO}_4 \) was the limiting reactant because it produced the smallest amount of potential product, \( 0.10 \text{ mol } \ ext{Ba}_3(\text{PO}_4)_2 \). Only when all reactants and their proportions are considered, you can accurately determine the maximum yield of a chemical reaction.

Mole Ratio

The mole ratio is a fundamental aspect of stoichiometry, playing a central role in determining the proportions of reactants and products in a chemical reaction. Derived from the coefficients of the balanced chemical equation, the mole ratio provides insight into how moles of one substance relate to moles of another substance in the reaction.
In the balanced equation \(3 \ ext{BaCl}_2 + 2 \ ext{Na}_3\text{PO}_4 \ ightarrow \ ext{Ba}_3(\text{PO}_4)_2 + 6 \ ext{NaCl} \), the mole ratio of \( ext{BaCl}_2 \) to \( ext{Ba}_3( ext{PO}_4)_2 \) is 3:1. This tells us that three moles of \( ext{BaCl}_2 \) are required to produce one mole of \( ext{Ba}_3( ext{PO}_4)_2 \). Similarly, the ratio of \( ext{Na}_3 ext{PO}_4 \) to \( ext{Ba}_3( ext{PO}_4)_2 \) is 2:1. Understanding these ratios is essential for calculating the amounts of reactants needed and predicting the amount of product formed.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, transform into new substances, known as products. This transformation occurs through the breaking of bonds in the reactants and the formation of new bonds in the products.
The balanced chemical equation represents this process using symbols. For example, in our exercise, the reaction between \( \ ext{BaCl}_2 \) and \( \ ext{Na}_3\text{PO}_4 \) forms \( \ ext{Ba}_3(\text{PO}_4)_2 \) and \( \ ext{NaCl} \), as shown in the balanced equation: \[3 \ ext{BaCl}_2 + 2 \ ext{Na}_3\text{PO}_4 \ ightarrow \ ext{Ba}_3(\text{PO}_4)_2 + 6 \ ext{NaCl} \]. Balancing an equation is key to stoichiometry because it ensures the conservation of mass; the number of each type of atom is the same on both sides of the equation. This reflects the real world, where matter is neither created nor destroyed, only transformed.