AIIMS 2017 Chemistry Limiting Reagents MCQ Question

In the reaction between BaCl₂ and Na₃PO₄, 2 moles of Na₃PO₄ produce 1 mole of Ba₃(PO₄)₂. Given that only 0.2 moles of Na₃PO₄ are available, the maximum amount of Ba₃(PO₄)₂ that can be formed is calculated as $\frac{0.2 \text{ moles Na₃PO₄}}{2} = 0.1 \text{ moles Ba₃(PO₄)₂}$. Other options are incorrect because they exceed the stoichiometric yield based on the limiting reagent, Na₃PO₄, which restricts the formation of Ba₃(PO₄)₂ to 0.1 moles.