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 fe2s3
b) 2.5g fe2s3
c) 3.5g fe2s3
d) 5.5g fe2s3

D) 5.5 g Fe₂S₃.

Explanation:

No. of moles of Fe filings (3.0 g) = mass /atomic mass = (3.0 g) / (55.845 g/mol) = 0.0537 mol.

No. of moles of yellow sulfur solid (2.5 g) = mass / molar mass = (2.5 g) / (256.52 g/mol) = 0.009746 mol.

The theoretical ratio of (Fe: S₈) is (16: 3) which is 5.333.

The actual ratio of the reactants (Fe:S₈) is (0.0537 mol: 0.009746 mol) which is 5.5.

Using cross multiplication:

3.0 moles of S₈ produces → 8.0 moles of Fe₂S₃, from the stichiometry,

0.009746 moles of S₈ produces → ??? moles of Fe₂S₃.

The mass of the the produced Fe₂S₃ =  no. of moles x molar mass = (0.026 mol)(207.9 g/mol) = 5.405 g ≅ 5.5 g.

it should have an internal chamber and two thermometers sticking out of the top. the internal chamber will also be closed off from the outside.