Question

0.5 moles bacl2 is mixed with 0.2 mole na3po4 find the maximum no of ba3(po4)2 that can be formed​

Answer 1

0.1 mole should be ans...........

3BaCl2 + 2Na3PO4  -------------> Ba3(PO4)2 + 6Nacl

                  0.5 mol           0.20 mol               x mol           

                First find which is the limiting reagent?

              0.5/3 = 1/6          0.2/2 = 1/10

                            1/10 < 1/6

              thus,  Na3PO4 is the limiting reagent.

               Now, 1/10 = x/1

                             x = 0.1 mol.