Question

When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3} .\) In a certain experiment, 20.00 \(\mathrm{g}\) iron metal was reacted with 11.20 \(\mathrm{g}\) oxygen gas. After the experiment, the iron was totally consumed, and 3.24 \(\mathrm{g}\) oxygen gas remained. Calculate the amounts of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed in this experiment.

Short answer

In this experiment, 19.05 g of FeO and 7.42 g of Fe2O3 are formed.

Step-by-step solution

Calculate the moles of iron and oxygen reacted

Initially, we are given 20.00 g of iron metal and 11.20 g of oxygen gas. After the reaction, 3.24 g of oxygen gas remains. To find the moles of iron and oxygen reacted, we will first find the moles of the initial amounts and then calculate the difference for oxygen. We will use the molar masses of the different elements. The molar mass of Fe = 55.84 g/mol The molar mass of O = 16.00 g/mol Moles of Fe_initial = \( \frac{20.00 g}{55.84 g/mol} \) = 0.3581 mol Moles of O_initial = \( \frac{11.20 g}{16.00 g/mol} \) = 0.7000 mol Moles of O_remaining = \( \frac{3.24 g}{16.00 g/mol} \) = 0.2025 mol Now, let's calculate the moles of reacted oxygen: Moles of O_reacted = Moles of O_initial - Moles of O_remaining = 0.7000 mol - 0.2025 mol = 0.4975 mol

Use stoichiometry to calculate the moles of FeO and Fe2O3 formed

For each mole of FeO formed, one mole of Fe and one mole of O react: Fe + O → FeO For each mole of Fe2O3 formed, two moles of Fe and three moles of O react: 4 Fe + 3 O2 → 2 Fe2O3 First, let's assume all the iron reacts to form FeO. Then, we can calculate the moles of FeO and the remaining moles of oxygen: Moles of FeO = Moles of Fe_reacted = 0.3581 mol Moles of O_left = Moles of O_reacted - Moles of FeO = 0.4975 mol - 0.3581 mol = 0.1394 mol Now, the remaining oxygen moles will react to form Fe2O3: Moles of Fe2O3 = \( \frac{1}{3} \) * Moles of O_left = \( \frac{1}{3} \) * 0.1394 mol = 0.04647 mol The moles of Fe required for forming 0.04647 mol of Fe2O3 can be calculated by multiplying by 2: Moles of Fe_required_for_Fe2O3 = 2 * Moles of Fe2O3 = 2 * 0.04647 mol = 0.09294 mol Now, let's adjust the moles of FeO formed since some iron was used for forming Fe2O3: Moles of FeO = Moles of FeO - Moles of Fe_required_for_Fe2O3 = 0.3581 mol - 0.09294 mol = 0.2652 mol

Calculate the mass of formed FeO and Fe2O3

Now that we have the moles of FeO and Fe2O3 formed, we can find the mass of each compound. The molar mass of FeO = 55.84 g/mol (Fe) + 16.00 g/mol (O) = 71.84 g/mol The molar mass of Fe2O3 = 2 * 55.84 g/mol (Fe) + 3 * 16.00 g/mol (O) = 159.69 g/mol Mass of FeO = Moles of FeO * Molar_mass_of_FeO = 0.2652 mol * 71.84 g/mol = 19.05 g Mass of Fe2O3 = Moles of Fe2O3 * Molar_mass_of_Fe2O3 = 0.04647 mol * 159.69 g/mol = 7.42 g Therefore, the amounts of FeO and Fe2O3 formed in this experiment are 19.05 g and 7.42 g, respectively.

Key concepts

Stoichiometry

Stoichiometry is the aspect of chemistry that focuses on the quantitative relationship between reactants and products in a chemical reaction. At its core, stoichiometry is about balancing these relationships to understand how much of each substance participates in a reaction.
In this particular reaction, stoichiometry helps us determine the amounts of iron oxide compounds formed when iron reacts with oxygen. It involves using the balanced chemical equations to relate the moles of reactants to those of the products.

Here's a simple breakdown of how stoichiometry is applied:

• First, we calculate the moles of each reactant using their respective molar masses.
• We then interpret the balanced chemical equations to understand what proportion of each reactant is needed to produce the products.
• Finally, stoichiometry allows us to translate these moles into actual masses, which is often what's needed to analyze reactions in a lab or real-world setting.

In our specific problem, stoichiometry enables us to determine that from reacting 20.00 g of iron entirely, a mixture of 19.05 g of FeO and 7.42 g of Fe2O3 is produced.

Iron Oxides

Iron oxides, which include compounds such as FeO and Fe2O3, are formed by the reaction of iron with oxygen. These oxides are common products in chemistry and play significant roles in various processes, from rusting to producing pigments.
In this context, iron oxides are the result of the limited oxygen supply reacting with iron, forming both FeO (iron(II) oxide) and Fe2O3 (iron(III) oxide).

Here's a basic comparison:

• FeO (Iron(II) oxide): Typically, when a lower amount of oxygen is available, FeO is formed. It involves a one-to-one ratio of reacting iron with oxygen.
• Fe2O3 (Iron(III) oxide): With relatively more oxygen, Fe2O3 forms, requiring a more significant amount of oxygen to react with iron in different proportions (4 Fe + 3 O2 → 2 Fe2O3).

Understanding these differences helps us appreciate the process mechanisms of forming different types of oxides and why it's crucial to control the conditions under which these reactions occur.

Limiting Reactant

In reactions, the limiting reactant is the substance that is completely consumed first and thus determines the maximum amount of product that can be formed. It is a crucial concept because it establishes the point at which the reaction stops.
For the problem at hand, identifying the limiting reactant is essential to calculate how much FeO and Fe2O3 form. In the given scenario, iron was fully consumed while oxygen remained, indicating that iron was the limiting reactant.

Here's how you can identify the limiting reactant:

• Compare the mole ratios of the reactants based on the balanced chemical equations.
• Calculate which reactant produces the least amount of product; this is the limiting step in forming the product.

By identifying iron as the limiting reactant, we calculated the exact quantities of the formed iron oxides, using all the iron available, which aligns with the data that oxygen was left over after the reaction.