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