Question

How many moles of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) can be formed when a mixture of \(0.36\) moles of aluminum and \(0.36\) moles of oxygen is ignited? Which substance and how much of it is in excess of that required? \(4 \mathrm{Al}+3 \mathrm{O}_{2} \rightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}\)

Short answer

In conclusion, 0.18 moles of aluminum oxide (\(\mathrm{Al}_{2}\mathrm{O}_{3}\)) can be formed when the given mixture of 0.36 moles aluminum and 0.36 moles oxygen is ignited. Oxygen is the excess reactant, with 0.09 moles remaining after the reaction.

Step-by-step solution

Identify the limiting reactant

To identify the limiting reactant, we first look at the balanced chemical equation: \[4 \mathrm{Al} + 3 \mathrm{O}_{2} \rightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}\] We are given 0.36 moles of both aluminum (\(\mathrm{Al}\)) and oxygen (\(\mathrm{O}_{2}\)). To determine which reactant will be used up first, we will compare the stoichiometric ratio between the reactants in the balanced chemical equation. In this case, the ratio is 4:3 (4 moles of aluminum react with 3 moles of oxygen). Next, we need to find the number of moles of oxygen required to react completely with the given moles of aluminum: \[\frac{4 \, \text{moles of Al}}{3 \, \text{moles of O}_{2}} = \frac{0.36 \, \text{moles}}{x \, \text{moles}}\] Solving for x, we have: \[x = \frac{3 \times 0.36}{4} = 0.27 \, \text{moles}\] Since we have more moles of oxygen (0.36 moles) than the 0.27 moles required to react completely with aluminum, aluminum is the limiting reactant.

Calculate the moles of the product formed

Now that we have identified aluminum as the limiting reactant, we can determine the moles of aluminum oxide (\(\mathrm{Al}_{2}\mathrm{O}_{3}\)) formed. Using the stoichiometry from the balanced chemical equation: \[2 \, \text{moles of Al}_{2} \mathrm{O}_{3} = \frac{4 \, \text{moles of Al}}{0.36 \, \text{moles of Al}}\] \[x = \frac{2 \times 0.36}{4} = 0.18 \, \text{moles}\] So, 0.18 moles of aluminum oxide can be formed from the reaction.

Determine the excess reactant and its remaining quantity

Oxygen is the excess reactant since aluminum was the limiting reactant. To determine the remaining moles of oxygen, we subtract the moles of oxygen that reacted with aluminum from the initial moles of oxygen: \[0.36 \, \text{moles of O}_{2} - 0.27 \, \text{moles of O}_{2} = 0.09 \, \text{moles of O}_{2}\] In conclusion, 0.18 moles of aluminum oxide (\(\mathrm{Al}_{2}\mathrm{O}_{3}\)) can be formed when the given mixture of 0.36 moles aluminum and 0.36 moles oxygen is ignited. Oxygen is the excess reactant, with 0.09 moles remaining after the reaction.

Key concepts

Stoichiometry

Stoichiometry is like the recipe book for chemical reactions. It helps us to relate the amounts of reactants and products using ratios derived from a balanced chemical equation. In our problem, stoichiometry allows us to determine how much aluminum oxide (\( \mathrm{Al}_{2} \mathrm{O}_{3} \)) can be formed from given amounts of aluminum and oxygen.

Let's break it down further:- **Reaction Ratios**: These ratios come from the balanced chemical equation, and they tell us how many moles of reactants are needed to produce a certain amount of product. In this case, the equation \( 4 \mathrm{Al} + 3 \mathrm{O}_{2} \rightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3} \) tells us that 4 moles of aluminum reacts with 3 moles of oxygen to yield 2 moles of aluminum oxide.

These stoichiometric ratios help us calculate not just the limiting reactant, but also the amount of product formed in the reaction. By relating all reactants and products in terms of moles, we can precisely understand how a chemical reaction progresses step by step.

Balanced Chemical Equation

At the heart of stoichiometry lies the balanced chemical equation. It's critical that an equation is balanced because it reflects the law of conservation of mass; matter is neither created nor destroyed.

Here's why the balanced equation is so important:- **Enrollment of Participating Atoms**: Each side of the equation must have the same number of atoms for every element involved. In our example, the equation \( 4 \mathrm{Al} + 3 \mathrm{O}_{2} \rightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3} \) shows 4 aluminum atoms and 6 oxygen atoms on both the reactant and product sides.- **Guiding Calculations**: The balanced equation guides stoichiometry. It's like setting proportions in a recipe, ensuring the correct mixture and amounts of ingredients for the perfect result.

This balance is crucial in determining which reactant limits the reaction and in calculating the amounts of each substance involved.

Mole Concept

The mole concept simplifies the counting of atoms and molecules using a standard unit, the mole, which corresponds to Avogadro's number (\(6.022 \times 10^{23}\) entities). This becomes the bridge between the atomic scale and our laboratory scale.

Key aspects of the mole concept in chemical reactions include:- **Mole-to-Mass Conversion**: Moles relate directly to the mass of substances via their molar masses, meaning we can shift from measuring in grams to counting in moles.- **Reactant Analysis**: In the given problem, each reactant's amount in moles (\(0.36\) moles of both aluminum and oxygen) is examined first to establish which one is a limiting factor. This simplifies our calculation of how many molecules partake in or result from a reaction.

Understanding the mole concept is fundamental for calculating how much of a product can be formed, deciding on limiting factors, and determining non-reacted excess substances.