Question

If 0.5 mole of BaCl₂ is mixed with 0.2 mole of Na₃PO₄ the maximum number of mole of Ba₃(PO₄)₂ that can be formed is 

1. 0.7
2. 0.5
3. 0.30
4. 0.10

Answer 1

Correct Answer - Option 4 : 0.10

Concept:

Stoichiometry of Chemical reactions:

• The study of chemical reactions and calculations related to the same is called Stoichiometry
• The coefficient used to balance the reaction is called the Stoichiometric coefficient.
• Stoichiometric ratios can be used to predict the moles of the products formed only if all the reactants are present in them.

Limiting reagents:

• The reactant, which is consumed first, limits the amount of product formed and is called the limiting reagent.
• In doing stoichiometric calculations, this aspect is also to be kept in mind.

Calculation:

Given: No.of moles of BaCl₂ = 0.5 mole; No.of moles of Na₃PO₄ = 0.2 mole

To find: No.of moles of Ba3(PO4)₂ =?

The balanced chemical reaction is as follows:

\(\begin{array}{*{20}{l}} {3{\text{ }}BaC{l_2}\; + {\text{ }}2\;N{a_3}P{O_4}\; \to {\text{ }}1{\text{ }}B{a_3}{{\left( {P{O_4}} \right)}_2}\; + {\text{ }}6{\text{ }}NaCl} \\ {\;{\text{ }}\;{\text{ }}\;\;\;\,\,\,{\mathbf{3}}\;{\text{ }}\;{\text{ }}:\;{\text{ }}\;{\text{ }}\;{\mathbf{2}}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;{\mathbf{1}}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;\,\,\,:\;{\text{ }}\;{\text{ }}\;{\text{ }}{\mathbf{6}}} \end{array}\)

According to the above reaction, 3 mol of BaCl₂ requires 2 mol of Na3PO₄ for the reaction.

Hence, for 0.5 mol of BaCl_2, the moles of Na3PO₄ required would be

\({\text{0}}{\text{.5}}\,{\text{mol}}\,{\text{BaC}}{{\text{l}}_{\text{2}}}{{ \times }}\frac{{{\text{2}}\,{\text{mol}}\,{\text{of}}\,{\text{N}}{{\text{a}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}}}{{{\text{3}}\,{\text{mol}}\,{\text{of}}\,{\text{BaC}}{{\text{l}}_{\text{2}}}\,}}{\text{ = 0}}{\text{.3}}\,{\text{mol}}\,{\text{N}}{{\text{a}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}\)

But we have only 0.2 mol Na3PO₄. Hence, Na3PO₄ is the limiting reagent in this case. So, the product is formed only from 0.2 mol Na3PO₄.

Since 0.2 mol Na3PO4 gives 1 mol Ba3(PO4)2 

\({\text{0}}{\text{.2}}\,{\text{mol}}\,{\text{N}}{{\text{a}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}{{ \times }}\frac{{{\text{1}}\,{\text{mol}}\,{\text{of}}\,{\text{B}}{{\text{a}}_{\text{3}}}{{{\text{(P}}{{\text{O}}_{\text{4}}}{\text{)}}}_{\text{2}}}\,}}{{{\text{2}}\,{\text{mol}}\,{\text{of}}\,{\text{N}}{{\text{a}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}}}{\text{ = }}\,{\text{0}}{\text{.}}\,{\text{1}}\,{\text{mol}}\,{\text{B}}{{\text{a}}_{\text{3}}}{{\text{(P}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{2}}}\)

Hence, the maximum no. of moles of Ba3(PO4)2 that can be formed is 0.1  mol.