Nickel (II) chloride, NiCl_2, reacts with sodium phosphate, Na_3PO_4, to produce

Question

Nickel (II) chloride, NiCl_2, reacts with sodium phosphate, Na_3PO_4, to produce a precipitate of nickel (II) phosphate, Ni_3(PO_4)_2, as shown the chemical equation below. 3 NiCl_2 (aq) + 2 Na_3PO_4 (aq) rightarrow Ni_3(PO_4)_2 (s) + 6 NaCl (aq) How many moles of nickel (II) chloride, NiCl_2, are required 3.31 moles of sodium chloride NaCl? Show how arrived at answer. How many moles of nickel (III) phosphate, Ni_3(PO_4)_2, can be made from 5.23 g sodium phosphate, Na_3PO_4? Show how you arrived at your answer.

Explanation / Answer

3NiCl2 + 2Na3PO4 ------------------------> Ni3(PO4)2 + 6NaCl

3                  2                                             1                    6

                                                                                       3.31

a)

3 moles NiCl2 --------------------> 6 moles NaCl

x mole NiCl2 ---------------------> 3.31 moles NaCl

moles of NiCl2 needed =x = 3 x 3.31 / 6 = 1.66

b)

Na3PO4 moles = 5.23 / 163.9 = 0.0319

2 mol Na3PO4 ----------------> 1 mole Ni3(PO4)2

0.0319 moles Na3PO4 -------------> 0.0319 x 1 / 2 = 0.0160 mol

moles of Ni3(PO4)2 = 0.0160

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