Question

A 100.0 ml portion of 0.250 M copper (ii) chloride is combined with 100.0 ml of 0.250 M potassium phosphate, resulting in the formation of a plate blue precipitate.
Part (A): what is the mass of the precipitate formed?
Part (B): how many moles of the excess reagent remain after the precipitation?
I need help with this question please.

Answer 1

Balanced equation:
3 CuCl₂(aq) + 2 K₃PO₄(aq) =====> Cu₃(PO₄)₂(s) + 6 KCl(aq)
Reaction type: double replacement

100.0 ml portion of 0.250 M copper (ii) chloride = 100 x 0.25 /1000 = 0.02 Moles

100.0 ml of 0.250 M potassium phosphate = 100 x 0.25 /1000 = 0.02 Moles

Acording to the balanced equation 3 equivalents of CuCl₂ needs for 2 equivalnet of K₃PO₄. Hence Limiting reagent is Copper chloride

For 0.02 Moles of Copper chloride product will be obtaind 0.00666 Moles

Hence 0.00666 Moles of Cu₃(PO₄)₂ = 0.006666 x 380.580 = 2.537 gm

2.537 gm of Cu₃(PO₄)₂ will be obtained

K₃PO₄ isExcess regent = 0.02 - 0.0133 = 0.00667 x 1000 /200 =0.03335 Moles

0.03335 Moles of K₃PO₄ Will be asExcess regent