Question

I am stuck at calculating mass of excess reactant used up in the reaction and mass of excess reactant left over after the reaction

2Na3PO4*12H2O + 3Ba Cl*2H2O --> Ba3(PO4)2 + 6NaCl + 30H2O

I got mass of precipitate : 0.063

Initial Mass of salt mixture 0.063

Moles of Ba3(PO4)2 precipitated: 1.04*10^-4

Moles of limiting reactant in the sal mixture : 3.14x10^-4

Mass of Limiting reactant in the salt Mixture :0.077 g

Mass of excess reactant in the salt mixture : 0.040 g

But I am stuck at calculate mass of excess reactant used up in the reaction and mass of excess reactant left over after the reaction

Help me plz

Answer 1

Moles of limiting reactanat is 3.14 x 10^-4

Mole sof product = 1.01 x 10^ -4 which is 1/3 of limiting reactanat

Hencelimiting reactanat is BaCl2.H2O since 3 BaCl2.H2O ives 1 Ba3(PO4)2

hence excess reactanat is Na3PO4.12H2O

as per reaction coeefiencts Na3PO4.12H2O moles needed in reaction = ( 2/3) BaCl2.H2O moles

                     = ( 2/3) x 3.14x10^-4 = 2.0933 x 10^- 4

mass of Na3PO4.12H2O used = moles used x molar mass = 2.0933x10^-4 x 380.12 = 0.0796 g

mass of excess reactanat left = 0.04 g

( pl check data again , I have did the procedure and used given values, I have taken mass of excess reactanat mass give 0.04 g as mass left , )