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). a) 1.5g fe b) 2.5g fe c) 3.5g fe d) 5.5g fe

1) Chemical equation

16Fe(s) + 3S8(s) ---> 8Fe2S3

2) Molar ratios:

16 mol Fe : 3 mole S8 : 8 mol Fe2S3

3) Convert masses in grams to number of moles

number of moles = mass in grams / molar mass

a) iron, Fe

mass = 3.0 g
atomic mass = 55.845 g/mol

=> number of moles of Fe = 3.0g / 55.845 g/mol = 0.0537 mol

b) Sulfur, S8

mass = 2.5 g
molar mass = 8*32.065 g/mol = 256.52 g/mol

=> number of moles of S8 = 2.5g / 256.52 g/mol = 0.009746 mol

4) Limiting reactant

Theoretical ratio                           actual ratio

16 mol Fe / 3 mol S8                 0.0537 mol Fe / 0.009746 mol S8

5.33                                               5.50

So, there is a little bit more Fe than the theoretical needed to react all the S8, which means the S8 is the limiting reactant.

5) Calculate the number of moles of iron (III) produced with 2.5 g (0.009746 moles) of S8

3moles S8 / 8 moles Fe2S3 = 0.009746 moles S8 / x

=> x = 0.009746 * 8 / 3 moles Fe2S3 = 0.026 moles Fe2S3

6) Convert 0.026 moles Fe2S3 into grams

mass in grams = number of moles * molar mass

molar mass of Fe2S3 = 207.9 g/mol

mass = 0.026 mol * 207.9 g/mol = 5.40 g

7) option D)

The answer is accuracy