Balance the reaction:
a FeS₂ + b O₂ => c Fe₂O₃ + d SO₂
Matching up the atoms on either side gives the system of equations
• Fe: a = 2c
• S: 2a = d
• O: 2b = 3c + 2d
We have 3 equations with 4 unknowns. Suppose we fix one of the variables, say a = 2. Then
2 = 2c ===> c = 1
2•2 = d ===> d = 4
2b = 3•1 + 2•4 ===> b = 11/2
and so the balanced reaction is
2 FeS₂ + 11/2 O₂ => Fe₂O₃ + 4 SO₂
or
4 FeS₂ + 11 O₂ => 2 Fe₂O₃ + 8 SO₂
Look up the molar masses of each component element:
• Fe: 55.845 g/mol
• S: 32.06 g/mol
• O: 15.999 g/mol
Then the molar masses of the reactants are
• FeS₂: 119.965 g/mol
• O₂: 31.998 g/mol
Convert the given masses to moles:
• (882 g FeS₂) (1/119.965 mol/g) ≈ 7.35 mol FeS₂
• (528 g O₂) (1/31.998 mol/g) ≈ 16.5 mol O₂
In the balanced reaction, 4 moles of FeS₂ combine with 11 moles of O₂, or equivalently 1 mol FeS₂ is needed for every 11/4 = 2.75 mol O₂. Now,
(16.5 mol O₂) / (2.75 mol O₂) = 6
while
(7.35 mol FeS₂) / (1 mol FeS₂) = 7.35
which means the reaction consumes all of the available O₂ and we're left with 7.35 - 6 = 1.35 mol FeS₂.
For every 11 mol O₂ among the reactants, 2 mol Fe₂O₃ are produced. Then 16.5 mol O₂ makes for
(11 mol O₂) / (2 mol Fe₂O₃) = (16.5 mol O₂) / x
===> x = 3.00 mol Fe₂O₃
which has a mass of
(3.00 mol Fe₂O₃) (160. g/mol) ≈ 479 g Fe₂O₃
