Question

If 0.5 mole of BaCl2 is mixed with 0.1 mole of Na3PO4, the maximum number of mole of Ba3(PO4)2 that can be formed is​

Answer 1

Answer:

0.05 moles

Step-by-step explanation:

The balanced chemical equation is as follows :

$3BaCl_2 +2Na_3PO_4 ---> Ba_3(PO_4)_2 +6NaCl$

=> 3 moles of BaCl₂ mixed with 2 moles of Na₃PO₄ forms 1 mole of Ba₃(PO₄)₂

 

According to question,

If 0.5 mole of BaCl₂ is mixed with 0.1 mole of Na₃PO₄, the maximum number of mole of Ba₃(PO₄)₂ that can be formed is​

=> 0.5 mole of BaCl₂ is mixed with 0.1 mole of Na₃PO₄ forms __ moles of Ba₃(PO₄)₂

⇒ number of moles of BaCl₂ required for 2 moles of Na₃PO₄ = 3

   number of moles of BaCl₂ required for 0.1 mole of Na₃PO₄

                                       = 0.1 × 3/2

                                       = 0.3/2

                                       = 0.15

                         

Given, number of moles of BaCl₂ reacted with 0.1 mole of Na₃PO₄ = 0.5

So, BaCl₂ is in excess

Limiting reagent = Na₃PO₄

=> 2 moles of Na₃PO₄ forms 1 mole of Ba₃(PO₄)₂

   0.1 mole of Na₃PO₄ forms ___ mole of Ba₃(PO₄)₂

                                        = 0.1 × 1/2

                                        = 0.1/2

                                        = 0.05

If 0.5 mole of BaCl2 is mixed with 0.1 mole of Na3PO4, the maximum number of mole of Ba3(PO4)2 that can be formed is​ 0.05 moles