2 Fe2S3 +902 - 2 Fe2O3 + 6 SO2 When 1.31 moles of Fe2S3 reacts with 22.8 moles of O2, how many moles of SO2 are formed

3.93 mol

Explanation:

The balanced equation is given as;

2 Fe₂S₃+ 9 O₂ --> 2 Fe₂O₃ + 6 SO₂

From the reaction;

2 mol of Fe₂S₃ reacts with 9 mol of O₂ to form 6 mol of 6 SO₂

This means;

1 mol of Fe₂S₃  requires 9/2 mol of O₂

Also,

1 mol of O₂ requires 1/9 mol of Fe₂S₃

In the question, we have;

1.31 moles of Fe2S3 and 22.8 moles of O2

The limiting reactant which determine how much of the product formed is; Fe₂S₃ because O₂ is in excess.

The relationship between Fe₂S₃ and SO₂ is;

2 mol Fe₂S₃ produces 6 mol of SO₂

1.31 mol of Fe₂S₃ would produce x mol ?

2 = 6

1.31 = x

x = 6 * 1.31 / 2 = 3.93 mol