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Chemistry - Mole Concept Question with Solution | TestHub

ChemistryMole ConceptLaws of Chemical CalculationsEasy2 minQB

ChemistryEasysingle choice

$3 BaCl_{2} + 2 Na_{3} PO_{4} ⟶ Ba_{3} (PO_{4})_{2} + 6 NaCl$ . Maximum amount of $Ba_{3} (PO_{4})_{2}$ formed when 2 moles each of $Na_{3} PO_{4}$ and $BaCl_{2}$ react is

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Answer:

Solution:

$2 B a C l_{2} + 2 N a_{3} P O_{4} ⟶ B a_{3} (P O_{4})_{2} + 6 N a Cl$

Here, $B a C l_{2}$ is the limiting reactant. The number of moles of $B a_{3} (P O_{4})_{2}$ formed is:

$\frac{1}{3} \times N o . o f m o l eso f B a C l_{2} = \frac{2}{3} m o l$

Stream:JEESubject:ChemistryTopic:Mole ConceptSubtopic:Laws of Chemical Calculations

⏱ 2mℹ️ Source: QB

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