How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 2.75 g of fe2s3 if the percent yield for the reaction is 65.0%? 3 na2s(aq) + 2 fecl3(aq) → fe2s3(s) + 6 nacl(aq)?

2FeCl3 + 3Na2S --> Fe2S3 + 6NaCl moles of Fe2S3 = 0.690 g / 207.9 g/mole = 0.00332 mole 0.00332 mole = X mole x 65.0% X mole = 0.00511 mole (at 100% yield) You require 2 mole FeCl3 to produce 1 mole of Fe2S3 (at 100% yield) moles of FeCl3 = 0.00511 x 2 = 0.0102 mole 0.0102 mole = 0.200 mole/liter x liters of FeCl3 liters of FeCl3 = 0.0511 (51.1 mL) ANSWER: 51.1 mL

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