Question

a 1.25g sample of ore containing iron pyrite (FeS2) was pulverized and ignited in air, converting the FeS2 to Fe2O3 and SO2(g). The Fe2O3 produced weighed 0.516g. What was the percent Fe in the ore

Answer 1

Answer:

28.9%

Explanation:

Let's consider the following balanced equation.

2 FeS₂ + 11/2 O₂ ⇒ Fe₂O₃ + 4 SO₂

We can establish the following relations:

• The molar mass of Fe₂O₃ is 159.6 g/mol
• 1 mole of Fe₂O₃ is produced per 2 moles of FeS₂
• 1 mole of Fe is in 1 mole of FeS₂
• The molar mass of Fe is 55.84 g/mol

The amount of Fe in the sample that produced 0.516 g of Fe₂O₃ is:

$0.516gFe_{2}O_{3}.\frac{1molFe_{2}O_{3}}{159.6gFe_{2}O_{3}} .\frac{2molFeS_{2}}{1molFe_{2}O_{3}} .\frac{1molFe}{1molFeS_{2}} .\frac{55.84gFe}{1molFe} =0.361gFe$

The percent of Fe in 1.25 g of the ore is:

$\frac{0.361g}{1.25g} .100\%=28.9\%$