Question

how many grams of iron III sulfide form when 62.0 mL of 0.135 M iron III chloride reacts with 45.0 mL of 0.285 M calcium sulfide. please show all steps neatly.

Answer 1

moles of FeCl3 = 62 x 0.135 / 1000 = 8.37 x 10^-3

moles of CaS = 45 x 0.285 / 1000 = 0.0128

2 FeCl3     + 3 CaS    ----------------> Fe2S3 + 3 CaCl2

    2                   3                                      1

8.37 x 10^-3    0.0128

here limiting reagent is FeCl3. so Fe2S3 formed based on this.

2 mol FeCl3   ------------> 1 mol Fe2S3

8.37 x 10^-3 mol   -----------> ??

mole of Fe2S3 = 8.37 x 10^-3 x 1 / 2 = 4.185 x 10^-3 mol

mass of Fe2S3 = 4.185 x 10^-3 x 207.9

mass of Fe2S3 = 0.870 g