I am stuck at calculating mass of excess reactant used up in the reaction and ma

ID: 949455 • Letter: I

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

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 , )

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