Predict the mass of iron (iii) sulfide produced when 3.0 g of iron filings react completely with 2.5 g of yellow sulfur solid, s8(s).

The complete balanced chemical reaction to this would be:

2 Fe  +  3/8 S8  -->  Fe2S3

First convert mass into number of moles.

moles Fe = 3 g / (55.845 g/mol) = 0.05372 mol Fe

moles S8 = 2.5 g / (256.52 g/mol) = 0.0097458 mol S8

Then we find the limiting reactant. The limiting reactant is the one which has lower (moles/coefficient) ratio.

Fe = 0.05372 / 2 = 0.02686

S8 = 0.0097458 / (3/8) = 0.02599

So since S8 has lower ratio, therefore it is the limiting reactant so we base our calculation from it. From the reaction, we get 1 mole of Fe2S3 for every 3/8 mol of S8, therefore:

moles Fe2S3 = 0.0097458 mol S8 * (1 mole Fe2S3 / 3/8 mol S8)

moles Fe2S3 = 0.02599 mol

The molar mass of Fe2S3 is 207.9 g/mol, so the mass is:

mass Fe2S3 = 0.02599 mol * 207.9 g/mol

mass Fe2S3 = 5.4 grams