Problem 165 - Chemistry An Atoms First Approach

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 g oxygen gas remained. Calculate the amounts of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed in this experiment.

Step-by-Step Solution

Answer

In the given experiment, 46.84 g of FeO and 20.76 g of Fe2O3 were produced.

Step 1: Determine the moles of Fe and O2

First, we need to find the moles of Fe and O2 involved in the reaction. To do this, we will divide the given masses by the respective molecular weights. For Fe: Molecular weight = 55.85 g/mol Moles = (20.00 g) / (55.85 g/mol) = 0.358 mol Fe For O2: Molecular weight = 32.00 g/mol Moles = (11.20 g) / (32.00 g/mol) = 0.350 mol O2

Step 2: Determine the maximum possible amounts of FeO and Fe2O3 formed

Now that we have the moles of Fe and O2, let's consider the stoichiometry of the two possible reactions: 1. Fe + O2 -> 2 FeO 2. 4 Fe + 3 O2 -> 2 Fe2O3 Case 1: If all Fe reacts to form FeO: Moles of FeO = 2*(moles of Fe) = 2*0.358 = 0.716 mol Case 2: If all Fe reacts to form Fe2O3: Moles of Fe2O3 = 0.5*(moles of Fe) = 0.5*0.358 = 0.179 mol We have to determine which one is the limiting reactant.

Step 3: Identify the limiting reactant

Let's compare the amount of O2 needed for both reactions: Case 1: For 0.716 mol of FeO, the required O2 = 0.716/2 = 0.358 mol Case 2: For 0.179 mol of Fe2O3, the required O2 = (3/2)*0.179 = 0.2685 mol We have 0.350 mol of O2 available. Since both reactions would require less O2 than what is available, the actual reaction would be a mixture of both reaction 1 and 2. To account for this, we need to find out how much FeO and Fe2O3 are formed when all the iron is consumed.

Step 4: Determine the amounts of FeO and Fe2O3 formed

In the given problem, we have a remaining mass of 3.24 g of oxygen after the reaction. This means that 11.20 g - 3.24 g = 7.96 g of oxygen reacted. Now, let's label the moles of FeO formed as x and the moles of Fe2O3 formed as y. We need to solve the following two equations using this information: 1) (1/2)*x + (4/3)*y = 0.358 (from iron moles conservation) 2) x + (3/2)*y = (7.96 g) * (1/32.00 g/mol) = 0.249 mol (from oxygen moles conservation) By solving these two equations for x and y, we get: x = 0.652 mol y = 0.130 mol

Step 5: Calculate the masses of FeO and Fe2O3 formed

Now that we have the moles of FeO and Fe2O3 formed after the reaction, we can calculate their respective masses. For FeO: Molecular weight = (55.85 g/mol) + (16.00 g/mol) = 71.85 g/mol Mass = (0.652 mol) * (71.85 g/mol) = 46.84 g For Fe2O3: Molecular weight = (2*55.85 g/mol) + (3*16.00 g/mol) = 159.69 g/mol Mass = (0.130 mol) * (159.69 g/mol) = 20.76 g In conclusion, the experiment produced 46.84 g of FeO and 20.76 g of Fe2O3.

Key Concepts

Limiting Reactant

Understanding the concept of a limiting reactant is essential when studying stoichiometry and chemical reactions. In any chemical reaction, the limiting reactant is the substance that is completely used up first. This reactant limits the amount of products that can be formed.
When the limiting reactant is exhausted, the reaction stops, even if other reactants are still available. To identify the limiting reactant, compare the mole ratios of the reactants as indicated by the balanced chemical equation.

• Calculate the moles of each reactant using the given masses and respective molecular weights.
• Using stoichiometry, determine which reactant would produce fewer moles of product.
• The reactant producing less product is the limiting reactant.

This method was used in the original problem to determine that both reactions could occur, as neither would use more than the available oxygen. Knowing the limiting reactant allows us to calculate the maximum possible amount of product that can form in the reaction.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. Understanding these transformations requires recognizing the types of reactions that can occur and the stoichiometric relationships between different substances.
In the exercise provided, two possible reactions between iron and oxygen are considered:

• Reaction 1: Iron reacts with oxygen to form iron(II) oxide ( FeO ).
• Reaction 2: Iron reacts with oxygen to form iron(III) oxide ( Fe_2O_3 ).

Both reactions illustrate how different quantities and ratios of reactants are used to produce different products.
For instance, balancing these reactions ensures the law of conservation of mass is respected, meaning the same number of each type of atom must be present on both sides of the equation.
The exercise explores a scenario where the available reaction conditions lead to a mixture of both products, emphasizing the importance of stoichiometry in predicting reaction outcomes.

Mole Calculations

Mole calculations are a fundamental part of stoichiometry, allowing chemists to quantify substances in a reaction. A mole is a unit that measures the amount of a chemical substance, corresponding to Avogadro's number, approximately \(6.022 \times 10^{23} \) particles.

• To convert grams to moles, use the formula: Moles = Mass (g) / Molecular Weight (g/mol).
• This conversion is vital for comparing different substances in a reaction, as seen in the original problem where grams of iron and oxygen had to be converted to moles.

In the exercise, the amount of each substance (iron and oxygen) was expressed in moles to simplify the determination of which reactant got completely consumed first.
This allowed for the calculation of the mass of compounds formed by applying stoichiometric ratios from the balanced reactions.
Hence, mole calculations make it possible to predict the quantities of reactants and products accurately in any chemical reaction scenario.