The Ml of 0.200 M FeCl3 needed to react with excess Na2s is calculated as follows
find the theoretical from % yield = actual yield / theoretical yield x100
let the theoretical yield be represented by y
that is = 0.345/y x100 = 65
0.345 x100= 65y
divide both side 65
y= 0.531 g
find moles =mass/molar mass
that is 0.531g/ 208 = 2.553 x10^-3 moles
write the equation for reaction
3Na2S + 2FeCl3 = Fe2S3 + 6NaCl
by use of mole ratio between FeCl3 to Fe2S3 which is 2:1 the moles of Fe2cl3
=2.553 x10^-3 x2 = 5.106 x10^-3 moles
volume = moles/molarity x1000
(5.106 x10^-3)/0.200 x1000 = 25.5 ml
