Question

How many moles of \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) are required to supply enough iron to prepare \(0.260 \mathrm{~mol} \mathrm{Fe}_{2} \mathrm{O}_{3}\) ? (Assume sufficient oxygen is available.)

Short answer

\(0.390 \text{ mol}\) of \(\mathrm{Fe}_{3}O_{4}\) is required.

Step-by-step solution

Determine the mole ratio

The balanced equation for the conversion of \(\mathrm{Fe}_{3}O_{4}\) to \(\mathrm{Fe}_{2}O_{3}\) is needed to determine the mole ratio. Since the direct equation is not provided, we base the conversion on iron alone, ignoring the oxygen since it's in excess. The conversion from \(\mathrm{Fe}_{3}O_{4}\) to \(\mathrm{Fe}_{2}O_{3}\) suggests that 3 moles of Fe from \(\mathrm{Fe}_{3}O_{4}\) would produce 2 moles of Fe in \(\mathrm{Fe}_{2}O_{3}\), giving a ratio of 3:2 for Fe.

Calculate the moles of \(\mathrm{Fe}_{3}O_{4}\) needed

Using the mole ratio from Step 1, we know that it takes 3 moles of Fe to make 2 moles of Fe in \(\mathrm{Fe}_{2}O_{3}\). Given that 0.260 moles of \(\mathrm{Fe}_{2}O_{3}\) are required, we calculate the necessary moles of \(\mathrm{Fe}_{3}O_{4}\) using the 3:2 ratio: \[ \frac{{\text{moles of }\mathrm{Fe}_{3}O_{4}}}{{\text{moles of }\mathrm{Fe}_{2}O_{3}}} = \frac{3}{2} \]. Solving for moles of \(\mathrm{Fe}_{3}O_{4}\), we get \[ \text{moles of }\mathrm{Fe}_{3}O_{4} = \frac{3}{2} \times \text{moles of }\mathrm{Fe}_{2}O_{3} \].

Plug in the given values

Now plug in the given value for moles of \(\mathrm{Fe}_{2}O_{3}\): \[ \text{moles of }\mathrm{Fe}_{3}O_{4} = \frac{3}{2} \times 0.260 \text{ mol} \].

Key concepts

Mole Ratio

Understanding the mole ratio in chemical reactions is essential for accurately determining the amounts of reactants needed and products formed. The mole ratio is derived from the coefficients of the balanced chemical equation, indicating the proportion of reactants and products involved in the reaction.

For example, let's consider a hypothetical reaction where reactant A converts to product B in a simple ratio. If the balanced equation is \[2A \rightarrow B\], it indicates that 2 moles of A will produce 1 mole of B. This 2:1 ratio is the mole ratio and is vital in calculations involving chemical reactions.

In the exercise provided, since sufficient oxygen is available, we focus on the iron atoms only. The mole ratio between \(\text{Fe}_{3}O_{4}\) and \(\text{Fe}_{2}O_{3}\) is derived from the ratio of iron atoms present in the compounds, which is 3:2. This tells us that for every 3 moles of iron in \(\text{Fe}_{3}O_{4}\), we get 2 moles of iron in \(\text{Fe}_{2}O_{3}\).

Chemical Reaction Balancing

Balancing chemical reactions is a foundational skill in chemistry that ensures the law of conservation of mass is respected. This law states that matter cannot be created or destroyed in a chemical reaction, which means the number of atoms for each element must be the same on both sides of the equation.

To balance a chemical reaction, we adjust the coefficients (the numbers in front of the chemical formulas) until the number of atoms of each element is equal on both sides of the equation. Remember, coefficients apply to all atoms in a compound, and only coefficients should be changed, not the subscripts in the chemical formulas, as doing so would change the compounds themselves.

Quick Tips:

• Start by balancing the atoms of elements that appear in only one reactant and one product.
• Balance polyatomic ions as a group if they appear unchanged on both sides of the equation.
• Finally, check that all atoms balance and that the total charge (if applicable) is the same on both sides.

In the context of the exercise, it's assumed that the reaction of \(\text{Fe}_{3}O_{4}\) to \(\text{Fe}_{2}O_{3}\) is already balanced, focusing solely on the iron atoms due to the excess of oxygen.

Molar Mass Calculations

The molar mass of a compound is the mass of one mole of that compound. It's calculated by summing the atomic masses of each element in the compound, multiplied by the number of atoms of that element in the formula. The atomic masses can be found on the periodic table and are typically expressed in grams per mole (g/mol).

For example, to find the molar mass of water (H₂O), we would add the molar mass of two hydrogen atoms (2 x 1.008 g/mol) to the molar mass of one oxygen atom (16.00 g/mol), resulting in approximately 18.02 g/mol for water.

Application to the Exercise:

Although not directly asked in the exercise, knowing how to calculate molar mass is crucial for subsequent steps, such as converting from moles to grams, which could be necessary if the question required the mass of \(\text{Fe}_{3}O_{4}\) instead of the number of moles. For the given exercise, molar mass calculations would enable you to find out how many grams of \(\text{Fe}_{3}O_{4}\) are required to react with the \(\text{Fe}_{2}O_{3}\). Such understanding emphasizes the interconnectedness of stoichiometry concepts in solving chemical problems.