how many millimeters of 0.200M FeCL3 are needed to react with an excess of Na2S to produce 1.38g of Fe2S3 if the percent yield for the reaction is 65.0%

1 answer

2FeCl3 + 3Na2S ==> Fe2S3 + 6NaCl

mols Fe2S3 = grams/molar mass = ? That's the mols you want if the reaction were 100% but it is only 65%; therefore, you want ? mol/0.65 = mols needed at 65% yield.
Using the coefficients in the balanced equation, convert mols Fe2S3 to mols
FeCl3.
Then M FeCl3 = mols FeCl3/L FeCl3. You know M and mols, solve for L and convert to mL.