Question

please help. me solve this question with work to understand

45) How many milliliters of 0.200 M Fe C13 are needed to react with an excess of Li2S to produce 0.345 g of Fe2S3 if the percent yield for the reaction is 65.0%? 3 Li2S(aq) + 2 Fe C13(aq) + Fe2S3(s) + 6 Li Cl(aq) A) 6.38 mL B) 10.8 mL C) 12.8 mL D) 25.5 mL E) 75.2 mL

Answer 1

Balanced equation:
3 Li2S + 2 FeCl3 ===> Fe2S3 + 6 LiCl
Reaction type: double replacement

Mass of Fe2S3 produced = 0.345 g

Moles of Fe2S3 produced = 0.345 g /207.88 g/mol =  0.001659 mole s

Moles of Fe2S3 produced for 100 % yield =  0.001659 mole x 100 % /65 % = 0.002552 Moles

Moles FeCl3 required =  0.005104 Moles

Concentration of FeCl3 = 0.2 M

Volume of FeCl3 required = 0.005104 x 1000 / 0.2 = 25.5 ml

Hence option D is correct answer