Question

How many liters of 3.38 M Na3PO4 are needed to produce 96.5 grams of NaCl? 2 Na3PO4(aq) + 3 CuCl2(aq) → Cu3(PO4)2(s) + 6 NaCl(aq)

Answer 1

number of moles of NaCl = 96.5g / 58.44 g.mol^-1 = 1.65 mole

from the balanced equation we can say that

6 mole of NaCl is produced by 2 mole of Na3PO4 so

1.65 mole of NaCl will be produced by

= 1.65 mole of NaCl *(2 mole of Na3PO4/6 mole of NaCl)

= 0.55 mole of Na3PO4

molarity of Na3PO4 = number of moles of Na3PO4 / volume of solution in L

3.38 = 0.55 / volume of solution in L

volume of solution in L = 0.55 / 3.38 = 0.163 L

Therefore, the volume of Na3PO4 =0.163 L