How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 0.345g of fe2s3 if the percent yield for the reaction is 65.0%?

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

Yes i think so not 100 percent sure though