Question

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. $$\mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)$$ If \(1.25 \mathrm{g}\) of iron is heated and placed in a bottle containing 0.0204 mol of oxygen gas, what mass of iron(III) oxide is produced?

Short answer

The limiting reactant is iron, and 1.79 grams of iron(III) oxide are produced when 1.25 g of iron reacts with 0.0204 mol of oxygen gas.

Step-by-step solution

Calculate the moles of iron and oxygen

First, we need to find the moles of iron and oxygen gas. The moles of iron can be calculated by dividing the mass of iron by its molar mass: Moles of iron = Mass of iron / Molar mass of iron Molar mass of iron (Fe) = 55.85 g/mol Moles of iron = 1.25 g / 55.85 g/mol = 0.0224 mol We are given the moles of oxygen gas, which is 0.0204 mol.

Determine the limiting reactant

Compare the molar ratio of the reactants with the stoichiometry of the reaction to determine the limiting reactant. The balanced chemical equation is: \(4 \mathrm{Fe}(s) + 3\mathrm{O}_{2}(g) \rightarrow 2\mathrm{Fe}_{2}\mathrm{O}_{3}(s)\) Divide the moles of each reactant by their stoichiometric coefficients: Moles of iron / 4 = 0.0224 mol / 4 = 0.0056 Moles of oxygen / 3 = 0.0204 mol / 3 = 0.0068 Iron has the smaller ratio (0.0056), so iron is the limiting reactant.

Calculate the moles of iron(III) oxide produced

Using the stoichiometry of the reaction, we can determine the moles of iron(III) oxide produced by multiplying the moles of the limiting reactant (iron) by the stoichiometric ratio of iron(III) oxide to iron: Moles of iron(III) oxide = Moles of iron × (2 mol iron(III) oxide / 4 mol iron) Moles of iron(III) oxide = 0.0224 mol × (2/4) = 0.0112 mol

Calculate the mass of iron(III) oxide produced

Finally, we will convert the moles of iron(III) oxide to mass using the molar mass of iron(III) oxide: Mass of iron(III) oxide = Moles of iron(III) oxide × Molar mass of iron(III) oxide The molar mass of iron(III) oxide is the sum of the molar masses of two iron atoms and three oxygen atoms: Molar mass of iron(III) oxide = 2 × 55.85 g/mol + 3 × 16 g/mol = 159.70 g/mol Mass of iron(III) oxide = 0.0112 mol × 159.70 g/mol = 1.79 g So, 1.79 grams of iron(III) oxide are produced.

Key concepts

Limiting Reactant

In any chemical reaction, the limiting reactant plays a crucial role as it determines the maximum amount of product that can be formed. It is the reactant that is entirely consumed first, thereby halting the reaction. Identifying the limiting reactant is essential for accurate stoichiometry calculations.

To find the limiting reactant, compare the mole ratio of each reactant with the stoichiometric coefficients from the balanced chemical equation:

• Divide the calculated moles of each reactant by their respective coefficients in the balanced equation.
• The reactant that has the smallest resultant value is the limiting reactant.

In the example provided, iron is the limiting reactant because it has a smaller ratio compared to oxygen when divided by their coefficients from the equation. This means that the amount of iron limits how much iron(III) oxide can be produced.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products through the breaking and forming of chemical bonds. In our example, iron and oxygen react to form iron(III) oxide, which is a classic example of a synthesis reaction.

The balanced chemical equation is vital because it shows the precise ratio in which the reactants combine. In the reaction:\[ 4 ext{Fe}(s) + 3 ext{O}_2(g) ightarrow 2 ext{Fe}_2 ext{O}_3(s) \]

• It indicates that 4 moles of iron react with 3 moles of oxygen to produce 2 moles of iron(III) oxide.
• Understanding this ratio helps in calculating the theoretical yield of products in mole or mass terms, based on the amounts of limiting reactant available.

Chemical equations not only show the substances involved but also highlight the conservation of mass, important for subsequent mole and mass calculations.

Mole Calculations

Mole calculations are fundamental in stoichiometry as they link the macroscopic world to the atomic scale. Knowing how to calculate the number of moles from given masses or volumes helps predict the amount of product formed in reactions.

The mole is a bridge between the atomic world and the macro world, defining amounts in terms of Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities per mole.

Here’s how mole calculations work in our example:

• To find the moles, divide the mass of a substance by its molar mass. For instance, the moles of iron were calculated by dividing its mass by its molar mass (55.85 g/mol).
• Similarly, if a reactant is in gaseous form, its moles might be given directly, as with the oxygen gas.

By calculating moles, students can determine how much of a product will form under given conditions, leading to practical applications in laboratory settings and industrial processes.