Question

Fe+O2=Fe2O2
If steel wool (iron) is heated until it glows and is placed in a bottle containing pure oxygen, the iron reacts spectacularly to produce iron(III) oxide.

If 1.28 g of iron is heated and placed in a bottle containing 0.0187 mol of oxygen gas, what mass of iron(III) oxide is produced?

Answer 1

Question #77397 - Chemistry - General Chemistry

$$\mathrm{Fe} + \mathrm{O}_2 = \mathrm{Fe}_2\mathrm{O}_3$$

If steel wool (iron) is heated until it glows and is placed in a bottle containing pure oxygen, the iron reacts spectacularly to produce iron(III) oxide.

If 1.28 g of iron is heated and placed in a bottle containing 0.0187 mol of oxygen gas, what mass of iron(III) oxide is produced?

Solution

$$4\mathrm{Fe} + 3\mathrm{O}_2 = 2\mathrm{Fe}_2\mathrm{O}_3$$

Mole Ratio 4:3:2

mass of Fe = 1.28 g

$$n(\mathrm{Fe}) = \frac{m(\mathrm{Fe})}{m_{\text{olar mass}}(\mathrm{Fe})}$$ $$M(\mathrm{Fe}) = 55.85\ \mathrm{g/mol}$$ $$n(\mathrm{Fe}) = \frac{1.28}{55.85} = 0.0229\ \mathrm{mol}$$ $$n(\mathrm{O}_2) = 0.0187\ \mathrm{mol}$$

First we need to find the limiting reagent.

n(Fe) needed = 0.0187 mol O₂ * 4 mol Fe/3 mol O₂ = 0.0249 mol

Therefore, Fe is limiting reagent.

Thus, we use the moles of Fe to continue with calculations.

$$n(\mathrm{Fe}_2\mathrm{O}_3) = 0.0229\ \mathrm{mol}\ (\mathrm{Fe}) * 2\ \mathrm{mol}\ (\mathrm{Fe}_2\mathrm{O}_3)/4\ \mathrm{mol}\ (\mathrm{Fe}) = 0.01145\ \mathrm{mol}$$ $$M(\mathrm{Fe}_2\mathrm{O}_3) = 159.70\ \mathrm{g/mol}$$ $$m(\mathrm{Fe}_2\mathrm{O}_3) = n(\mathrm{Fe}_2\mathrm{O}_3) * M(\mathrm{Fe}_2\mathrm{O}_3) = 0.01145 * 159.70 = 1.829\ \mathrm{g}$$

Answer

1.829 g iron(III) oxide is produced.

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