Problem 79 - Chemistry Matter and Change

Question

Iron Production Iron is obtained commercially by the reaction of hematite \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) with carbon monoxide. How many grams of iron is produced when 25.0 \(\mathrm{mol}\) of hematite reacts with 30.0 \(\mathrm{mol}\) of carbon monoxide? \begin{equation} \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g}) \end{equation}

Step-by-Step Solution

Answer

1117.0 grams of iron are produced.

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for the reaction is given as: \[\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})\] This equation indicates that one mole of \(\mathrm{Fe}_2 \mathrm{O}_3\) reacts with three moles of \(\mathrm{CO}\) to produce two moles of \(\mathrm{Fe}\).

Step 2: Find the Limiting Reactant

Given: 25.0 moles of \(\mathrm{Fe}_2 \mathrm{O}_3\) and 30.0 moles of \(\mathrm{CO}\). The stoichiometric ratio is 1 mole \(\mathrm{Fe}_2 \mathrm{O}_3\) : 3 moles \(\mathrm{CO}\). Calculate moles of \(\mathrm{CO}\) required for 25.0 moles of \(\mathrm{Fe}_2 \mathrm{O}_3\): \[25.0 \text{ mol } \mathrm{Fe}_2 \mathrm{O}_3 \times \frac{3 \text{ mol } \mathrm{CO}}{1 \text{ mol } \mathrm{Fe}_2 \mathrm{O}_3} = 75.0 \text{ mol } \mathrm{CO}\] Since only 30.0 moles of \(\mathrm{CO}\) are available, \(\mathrm{CO}\) is the limiting reactant.

Step 3: Calculate Moles of Iron Produced

Using the stoichiometric ratio from the balanced equation, \(\text{3 mol } \mathrm{CO} : \text{2 mol } \mathrm{Fe}\), calculate the moles of \(\mathrm{Fe}\) produced from 30.0 moles of \(\mathrm{CO}\): \[30.0 \text{ mol } \mathrm{CO} \times \frac{2 \text{ mol } \mathrm{Fe}}{3 \text{ mol } \mathrm{CO}} = 20.0 \text{ mol } \mathrm{Fe}\]

Step 4: Convert Moles of Iron to Grams

The molar mass of iron (\(\mathrm{Fe}\)) is approximately 55.85 g/mol. Convert moles of iron to grams: \[20.0 \text{ mol } \mathrm{Fe} \times 55.85 \text{ g/mol} = 1117.0 \text{ g } \mathrm{Fe}\] Thus, 1117.0 grams of iron are produced.

Key Concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It involves using the coefficients from the balanced chemical equation to determine the ratios of reactants and products.
This helps us understand how much of a reactant is needed to produce a desired amount of product or how much product can be expected from a certain amount of reactant.In the given problem of iron production, stoichiometry allows us to calculate how many moles of carbon monoxide are needed to react completely with hematite (\(\mathrm{Fe}_{2}\mathrm{O}_{3}\)) to produce iron. This calculation ensures that all elements are conserved, and no atoms are lost during the reaction.
So, stoichiometry essentially acts as a recipe that tells us the exact proportions of ingredients needed.

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is completely consumed first and thus limits the amount of product that can be formed. It essentially determines when the reaction will stop.
Identifying the limiting reactant involves comparing the mole ratio of the given reactants with the ratios from the balanced chemical equation.For the reaction between hematite and carbon monoxide, we are given 25.0 moles of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) and 30.0 moles of \(\mathrm{CO}\). Using stoichiometry, we need 75.0 moles of \(\mathrm{CO}\) to react with the given 25.0 moles of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\).
Since we only have 30.0 moles of \(\mathrm{CO}\), it limits the amount of iron produced, making \(\mathrm{CO}\) the limiting reactant in this scenario.

Mole-to-Mass Conversion

Converting moles to mass is a fundamental skill in chemistry that enables us to translate a calculated amount of a substance from the molecular scale to a tangible mass that we can measure. This involves using the molar mass of a substance, which is the mass of one mole of that substance expressed in g/mol.In the context of our problem, once we've determined the limiting reactant and calculated the moles of iron (\(\mathrm{Fe}\)) produced, the next step is to convert these moles into grams. To do this, we multiply the moles of iron produced by its molar mass.The molar mass of iron is approximately 55.85 g/mol. Therefore, if we have 20.0 moles of iron, it translates to \[20.0 \text{ mol } \mathrm{Fe} \times 55.85 \text{ g/mol} = 1117.0 \text{ g } \mathrm{Fe}\].
This conversion allows chemists to measure and use specific quantities of substances in practical applications.

Balanced Chemical Equation

A balanced chemical equation is an equation that shows the reactants and products of a chemical reaction with the same number of atoms of each element on both sides of the equation. This ensures that mass is conserved according to the law of conservation of mass.To balance the equation for the reaction between hematite and carbon monoxide, we first write the unbalanced equation and then adjust the coefficients to ensure that the same number of each type of atom appears on both sides of the equation. The balanced equation given in this exercise is:\[\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})\].
This equation tells us that one mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacts with three moles of \(\mathrm{CO}\) to produce two moles of \(\mathrm{Fe}\) and three moles of \(\mathrm{CO}_{2}\). Each component is systematically adjusted to reflect reality at a molecular level.
Remember: Balancing chemical equations is crucial for accurate calculations in stoichiometry and real-world chemical applications.