Question

When aqueous solutions of K3PO4 and Ba(NO3)2 are combined, Ba3(PO4)2 precipitates. Calculate the mass, in grams, of the Ba3(PO4)2 produced when 1.84 mL of 0.129 M Ba(NO3)2 and 3.76 mL of 0.712 M K3PO4 are mixed. Calculate the mass to 3 significant figures.

Answer 1

This reaction can only occur at very high temperatures, if the reaction occured at high temperature then

Number of moles of Ba(NO3)2 = 1.89/1000 * 0.129 = 0.00024831 moles

Number of moles of K3PO4 = 3.76/1000 * 0.712 = 0.0026712 moles

Ba(NO3)2 will be the limiting reagenet

3 moles of Ba(NO3)2 produces 1 mole of precipitate

Moles of Ba3(PO4)2 = 0.00024831/3 moles = 0.00008277 moles

Molar mass of Precipitate (Ba3(PO4)2) = 601.9237 gms

Mass of Ba3(PO4)2 = 0.00008277 * 601.9237 = 0.0498