Problem 73 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(0.50\) mole of \(\mathrm{BaCl}_{2}\) is mixed with \(0.20\) mole 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 (a) \(0.10\) (b) \(0.20\) (c) \(0.30\) (d) \(0.40\)

Step-by-Step Solution

Answer

(a) 0.10

Step 1: Write the balanced chemical equation

The reaction between barium chloride (\(\mathrm{BaCl}_2\)) and sodium phosphate (\(\mathrm{Na}_3\mathrm{PO}_4\)) can be written as follows: \[3\mathrm{BaCl}_2 + 2\mathrm{Na}_3\mathrm{PO}_4 \rightarrow \mathrm{Ba}_3(\mathrm{PO}_4)_2 + 6\mathrm{NaCl}\] This equation shows that 3 moles of \(\mathrm{BaCl}_2\) react with 2 moles of \(\mathrm{Na}_3\mathrm{PO}_4\) to form 1 mole of \(\mathrm{Ba}_3(\mathrm{PO}_4)_2\).

Step 2: Determine the limiting reactant

Given: 0.50 moles of \(\mathrm{BaCl}_2\) and 0.20 moles of \(\mathrm{Na}_3\mathrm{PO}_4\). From the balanced equation, the mole ratio required is 3:2. Let's compare the given moles with this ratio: - For \(\mathrm{BaCl}_2\): \(\frac{0.50}{3} = 0.167\)- For \(\mathrm{Na}_3\mathrm{PO}_4\): \(\frac{0.20}{2} = 0.10\)Since \(\mathrm{Na}_3\mathrm{PO}_4\) has the lesser value in the ratio (0.10), it is the limiting reactant.

Step 3: Calculate maximum moles of product

From the limiting reactant: 0.20 moles of \(\mathrm{Na}_3\mathrm{PO}_4\) when fully reacted as per the equation forms \(\frac{0.20}{2} = 0.10\) moles of \(\mathrm{Ba}_3(\mathrm{PO}_4)_2\). Thus, the maximum number of moles of \(\mathrm{Ba}_3(\mathrm{PO}_4)_2\) that can be formed is 0.10.

Key Concepts

Limiting Reactant

When reacting two or more substances, the limiting reactant is the one that will be completely used up first in the reaction. It determines the maximum amount of product that can be formed. Imagine making sandwiches: if you have 8 slices of bread and 5 slices of cheese, you can only make 4 cheese sandwiches. Here, the cheese is the limiting ingredient because you run out of it first.
In the given exercise, we are looking at a similar idea but with chemicals. We have 0.50 moles of \(\mathrm{BaCl}_2\) and 0.20 moles of \(\mathrm{Na}_3\mathrm{PO}_4\). By checking the balanced chemical equation, which tells us we need 3 moles of \(\mathrm{BaCl}_2\) for every 2 moles of \(\mathrm{Na}_3\mathrm{PO}_4\), we can calculate the ratio of these reactants.

• For \(\mathrm{BaCl}_2\): \(\frac{0.50}{3} \approx 0.167\)
• For \(\mathrm{Na}_3\mathrm{PO}_4\): \(\frac{0.20}{2} = 0.10\)

Here, \(\mathrm{Na}_3\mathrm{PO}_4\) is the limiting reactant because 0.10 is the smaller value. Understanding which is the limiting reactant lets us calculate how much product can form.

Balanced Chemical Equation

A balanced chemical equation is like a recipe providing the exact proportions of ingredients needed for a chemical reaction. This ensures that the mass and number of atoms are conserved during the reaction. For the reaction between \(\mathrm{BaCl}_2\) and \(\mathrm{Na}_3\mathrm{PO}_4\), the balanced equation is given as:\[3\mathrm{BaCl}_2 + 2\mathrm{Na}_3\mathrm{PO}_4 \rightarrow \mathrm{Ba}_3(\mathrm{PO}_4)_2 + 6\mathrm{NaCl}\]This setup tells us that 3 moles of \(\mathrm{BaCl}_2\) are required to react fully with 2 moles of \(\mathrm{Na}_3\mathrm{PO}_4\). The equation also indicates we will produce 1 mole of \(\mathrm{Ba}_3(\mathrm{PO}_4)_2\) and 6 moles of \(\mathrm{NaCl}\).
Balancing the equation ensures that we have the same number of each atom on both sides, signifying that no atoms are lost - they are simply rearranged.

• 3 Ba atoms on each side
• 6 Cl atoms on each side
• 2 \(\mathrm{PO}_4\) groups on each side
• 6 Na atoms on each side

Being able to interpret these equations is crucial in predicting the outcome of the reaction and maximizing the product.

Mole Concept

The mole is a fundamental unit in chemistry, used to express amounts of a chemical substance. One mole equals \(6.022 \times 10^{23}\) entities, which could be atoms, molecules, or ions, making it a vital tool in chemical calculations. Setting a common scale helps chemists communicate and calculate chemical quantities over widely varying scales.
In the context of the exercise, we used moles to understand the chemical ratio involved in the balanced equation. With this, we were able to identify the limiting reactant and determine how much \(\mathrm{Ba}_3(\mathrm{PO}_4)_2\) could be produced.

Using the Mole Concept to Calculate Product

To find out the possible product formed, we start with the moles of the limiting reactant. Here, 0.20 moles of \(\mathrm{Na}_3\mathrm{PO}_4\),

• From the equation, 2 moles of \(\mathrm{Na}_3\mathrm{PO}_4\) produces 1 mole of \(\mathrm{Ba}_3(\mathrm{PO}_4)_2\).
• So \(\frac{0.20}{2}=0.10\) moles of \(\mathrm{Ba}_3(\mathrm{PO}_4)_2\) is produced.

The mole concept allows us to convert these quantities into understandable and comparable units, highlighting its essential role in stoichiometry.