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

Question

If \(0.50\) mole of \(\mathrm{BaCl}_{2}\) is mixed with \(0.20 \mathrm{~mole}\) 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

0.10 moles of \(\mathrm{Ba}_3(\mathrm{PO}_4)_2\) can be formed.

Step 1: Write the Chemical Reaction Equation

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

Step 2: Determine Mole Ratios

From the chemical equation, the mole ratio of \(\mathrm{BaCl}_2\) to \(\mathrm{Na}_3\mathrm{PO}_4\) is 3:2. This ratio tells us the proportions in which the reactants react to form the product.

Step 3: Identify the Limiting Reactant

Calculate the number of moles of both reactants available. We have 0.50 mole of \(\mathrm{BaCl}_2\) and 0.20 mole of \(\mathrm{Na}_3\mathrm{PO}_4\). According to the stoichiometry from the equation, 3 moles of \(\mathrm{BaCl}_2\) are needed for every 2 moles of \(\mathrm{Na}_3\mathrm{PO}_4\). Therefore, to react with 0.20 mole of \(\mathrm{Na}_3\mathrm{PO}_4\), we would require \(0.20 \times \frac{3}{2} = 0.30\) moles of \(\mathrm{BaCl}_2\). Since we have more than 0.30 moles (specifically 0.50 moles) of \(\mathrm{BaCl}_2\), \(\mathrm{Na}_3\mathrm{PO}_4\) is the limiting reactant.

Step 4: Calculate Maximum Product Formation

From the stoichiometric equation, 2 moles of \(\mathrm{Na}_3\mathrm{PO}_4\) produce 1 mole of \(\mathrm{Ba}_3(\mathrm{PO}_4)_2\). Hence, 0.20 mole of \(\mathrm{Na}_3\mathrm{PO}_4\) will produce \(0.20 \times \frac{1}{2} = 0.10\) moles of \(\mathrm{Ba}_3(\mathrm{PO}_4)_2\).

Key Concepts

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that gets completely used up first, stopping the reaction from proceeding any further because there are no more molecules of this reactant to combine. Determining the limiting reactant requires you to compare the mole ratio of the amount of reactants you start with to the mole ratio required by the balanced equation.
As shown in the problem, with our initial 0.50 moles of \( \mathrm{BaCl}_2 \) and 0.20 moles of \( \mathrm{Na}_3\mathrm{PO}_4 \), there is enough \( \mathrm{BaCl}_2 \) to react with all of the \( \mathrm{Na}_3\mathrm{PO}_4 \) given the mole ratio. \( \mathrm{Na}_3\mathrm{PO}_4 \) can run out first because you need 3 moles of \( \mathrm{BaCl}_2 \) for 2 moles of \( \mathrm{Na}_3\mathrm{PO}_4 \), making \( \mathrm{Na}_3\mathrm{PO}_4 \) the limiting reactant in this case.

• Identifying the limiting reactant is crucial as it determines the maximum amount of product that can be formed.
• To do this, calculate how many moles of each reactant you'll need to use each up fully with the other reactant.
• The reactant for which you need more moles than you have is the limiting reactant.

Mole Ratios

Mole ratios are fundamental in stoichiometry and provide information about the proportions of reactants and products in a chemical reaction. They are derived from the coefficients in the balanced chemical equation.
In the given reaction:
\[ 3\,\mathrm{BaCl}_2 + 2\,\mathrm{Na}_3\mathrm{PO}_4 \rightarrow \mathrm{Ba}_3(\mathrm{PO}_4)_2 + 6\,\mathrm{NaCl} \]
The coefficients indicate the mole ratios: 3 moles of \( \mathrm{BaCl}_2 \) react with 2 moles of \( \mathrm{Na}_3\mathrm{PO}_4 \) to produce 1 mole of \( \mathrm{Ba}_3(\mathrm{PO}_4)_2 \), and 6 moles of \( \mathrm{NaCl} \).

• Using these ratios, you can calculate how much of one substance is needed to react with a given amount of another substance.
• This is especially important when identifying the limiting reactant, as it tells you exactly how much you need to "use it up."
• In our example, finding the limiting reactant involved using the \( \frac{3}{2} \) ratio between \( \mathrm{BaCl}_2 \) and \( \mathrm{Na}_3\mathrm{PO}_4 \), which indicates the proportion needed for the reaction to complete.

Balanced Chemical Equation

A balanced chemical equation adheres to the law of conservation of mass and shows equal numbers of each type of atom on both sides of the equation. This is essential for accurately representing chemical reactions, as it allows scientists to predict the quantities of products that will form from given quantities of reactants.
Balancing equations involves:

• Counting the number of each type of atom on both sides of an equation.
• Adding coefficients to molecules to ensure the number of atoms stays the same.

For example, the equation:
\[ 3\,\mathrm{BaCl}_2 + 2\,\mathrm{Na}_3\mathrm{PO}_4 \rightarrow \mathrm{Ba}_3(\mathrm{PO}_4)_2 + 6\,\mathrm{NaCl} \]
is balanced because:

• There are 3 barium, 14 chloride, 2 phosphorus, and 8 oxygen atoms on both the reactant and product sides.

This balancing ensures that when reactants transform into products, no atoms are lost or created, just rearranged.
It allows chemists to use stoichiometry to calculate actual amounts needed and predict product outcomes accurately.