Question

23.1 mL of 2.10 M Ba(OH)₂ is added to 73.1 mL of 0.813 M H₃PO₄. What mass of Ba₃(PO₄)₂ (s) is formed from this reaction? Ba₃(PO₄)₂ (s) is from this reaction? Ba₃(PO₄)₂ is barium phosphate and is insoluble in water.

(a) 9.73 g

(b) 35.7 g

(c) 14.6 g

(d) 17.9 g

Answer 1

The reaction between Ba(OH)2 and H3PO4 is

3Ba(OH)2+2 H3PO4---> Ba3(PO4)₂ +6H2O

molar ratio of Ba(OH)2 : H3PO4= 3:2 =1.5:1

moles of Ba(OH)2 in 23.1ml of 2.1M= 2.1*23.1/1000 moles=0.0485 moles

moles of H3PO4 in 73.1ml of 0.813M= 0.813*73.1/1000=0.05943

Molar ratio (actual )= 0.0485: 0.05943 = 1,23 :1

So limiting reactants is Ba(OH)2 since it is required is 1.23 against 1.5

So moles of Ba3(PO4)2 formed is limited by Ba(OH)2 moles which are 0.0485

moles of Ba3(PO4)2 formed= 0.0485/3= 0.0162 moles

molar mas of Ba3(PO4)2= 602

Mass of Ba3(PO4)2 formed =0.0162*602= 9.73 gm ( A is correct)