In a reaction vessel 0.600 mole of Ba(NO3)2(aq) and 0.300 molly of H3 PO4(aq) are combined with dionized water to a final volume of 2.00L.

Question

3Ba(NO3)2 +2H3PO4=Ba3(PO4)2+6HNO3

1.Calculate the mass of Ba3(PO4)2 formed. 

2. What is the concentration in Molarity of NO3(aq). After the reaction reaches completion .? 

 

Answer 1

3Ba(NO3)2 +2H3PO4=Ba3(PO4)2+6HNO3

First find limiting reactant:

— 0.600 mole 3Ba(NO3)2/3 = 0.200 mole

— 0.300 mole H3PO4/2 = 0.150 mole

— Limiting reactant is H3PO4

1.  Mass Ba3(PO4)2 = 0.15 mol H3PO4 x 1 mol Ba3(PO4)2/2 mole H3PO4 = 0.075 mol x 602g/mol = 45.2 g

 

2.  0.15 H3PO4 x 6HNO3/2 H3PO4 = 0.45 mol HNO3

M = 0.45 mol/2.00 L = 0.225 M NO3^-