Question

What volume, in mL, of 0.109 M NiCl2 solution is required to form 78.8 grams of precipitate? Assume excess Na3PO4.

3NiCl2 (aq) + 2Na3PO4 (aq) = Ni3(PO4)2 (s) + 6NaCl (aq)

thank you!

Answer 1

Mass of precipitate is 78.8 grams.

Ni3(PO4)2

Molar mass of precipitate is 366 g/mol.

Ni3(PO4)2

Number of moles of

78.8 g Ni3(PO4)2 = 366 g/mol = 0.215 mol

3 mol Nicl 0.215 mol Ni3(PO4)2 x 1 mol Ni3(PO4)2 -= 0.646 mol Nicly

Volume of

0.646 mol NiCl2 = 0.109 moi = 5.93 L

Convert unit from L to mL

1000 ml 5.93 LX = 593 mL 1L