Question

What volume, in L, of 0.169 M NiCl2 solution is required to produce 73.4 g of precipitate.

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

Answer 1

we have the Balanced chemical equation as:

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

Molar mass of Ni3(PO4)2 = 3*MM(Ni) + 2*MM(P) + 8*MM(O)

= 3*58.69 + 2*30.97 + 8*16.0

= 366.01 g/mol

mass of Ni3(PO4)2 = 73.4 g

we have below equation to be used:

number of mol of Ni3(PO4)2,

n = mass of Ni3(PO4)2/molar mass of Ni3(PO4)2

=(73.4 g)/(366.01 g/mol)

= 0.2005 mol

From balanced chemical reaction, we see that

when 1 mol of Ni3(PO4)2 forms, 3 mol of NiCl2 reacts

mol of NiCl2 reacted = (3/1)* moles of Ni3(PO4)2

= (3/1)*0.2005

= 0.6016 mol

This is number of moles of NiCl2

we have below equation to be used:

M = number of mol / volume in L

0.169 = 0.6016/ volume in L

volume = 3.56 L

Answer: 3.56 L