Question

Consider the reaction:

2 Na₃PO₄(aq) + 3 Cu(NO₃)₂(aq) → Cu₃(PO₄)₂(s) + 6 NaNO₃(aq)

What mass of Cu₃(PO₄)₂ can be formed when 71.2 mL of a 0.890 M solution of Na₃PO₄ is mixed with 28.8 mL of a 0.140 M solution of Cu(NO₃)₂?

___________ g

Answer 1

find out the no o fmoles of Na3PO4 and Cu(NO3)2 using the formula

no of moles = Molarity x volume in liters

no of moles of Na3PO4 = 0.890 M x 0.0712 L = 0.063 moles

no of moles of Cu(NO3)2 = 0.140M x 0.0288L = 0.004 moles

now look at the balanced equation

2 Na3PO4(aq) + 3 Cu(NO3)2(aq) → Cu3(PO4)2(s) + 6 NaNO3(aq)

for getting one mole of Cu3(PO4)2(s) we need 2 moles of Na3PO4 amd 3 moles  Cu(NO3)2

but here less no of Cu(NO3)2 is there means limiting reagent is Cu(NO3)2

so we have to calculate mass of Cu3(PO4)2(s) with respect to limiting agent

fom the balanced equation

from one mole of Cu(NO3)2 1/3 moles of Cu3(PO4)2(s) is forming

from 0.004 moles of Cu(NO3)2 we will get 1/3 x 0.004 = 0.00134 moles of Cu3(PO4)2

now weight of Cu3(PO4)2 = moles of Cu3(PO4)2 x molar mass of Cu3(PO4)2

= 0.00134moles x 380.6 g/mole

= 0.511 grams