Question

Consider the following reaction: $$4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)$$ It takes 2.00 L of pure oxygen gas at STP to react completely with a certain sample of aluminum. What is the mass of aluminum reacted?

Short answer

The mass of aluminum that reacted with 2.00 L of pure oxygen gas at STP is 3.214 g.

Step-by-step solution

Determine the number of moles of oxygen gas

Using the volume of oxygen gas and the conditions at STP (Standard Temperature and Pressure), we can find the number of moles of oxygen gas. Recall that at STP, 1 mole of any gas occupies a volume of 22.4 L. Thus, we can find the moles of oxygen gas using the following formula: Moles of O₂ = Volume of O₂ / Volume occupied by 1 mole of O₂ at STP Moles of O₂ = 2.00 L / 22.4 L/mol Moles of O₂ = 0.0893 mol

Determine the number of moles of aluminum

Now that we know the number of moles of oxygen gas, we can use the stoichiometry of the balanced equation to find the number of moles of aluminum required for the reaction: 4 Al + 3 O₂ → 2 Al₂O₃ From the balanced equation, we can see that 3 moles of O₂ react with 4 moles of Al. Using this ratio, we can calculate the number of moles of aluminum: Moles of Al = (moles of O₂ × moles of Al) / moles of O₂ Moles of Al = (0.0893 mol × 4 mol) / 3 mol Moles of Al = 0.1191 mol

Convert the number of moles of aluminum to mass

Now that we have found the number of moles of aluminum, we can convert it to mass using the molar mass of aluminum (26.98 g/mol): Mass of Al = moles of Al × molar mass of Al Mass of Al = 0.1191 mol × 26.98 g/mol Mass of Al = 3.214 g Therefore, the mass of aluminum that reacted with oxygen gas is 3.214 g.

Key concepts

Chemical Reactions

A chemical reaction is a process where reactants transform into products due to the breaking and forming of chemical bonds.
The equation provided in the exercise demonstrates one such reaction:

• 4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

This equation tells us that aluminum (Al) reacts with oxygen (O₂) to form aluminum oxide (Al₂O₃).
Each component in the equation has coefficients indicating the number of moles required. In our case:

• 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide.

Using these stoichiometric relationships, we can compute quantities of reactants or products when some are already known.
Understanding these relationships is crucial for solving problems related to chemical reactions and predicting the outcomes of reactions.

Moles and Molar Mass

The concept of a mole is foundational in chemistry, allowing us to quantify the amount of a substance.
A mole is Avogadro's number of particles, atoms, or molecules (approximately 6.022 × 10²³).
Molar mass, on the other hand, is the mass of one mole of a substance, usually expressed in g/mol.

• In the reaction, aluminum has a molar mass of 26.98 g/mol.

Determining the moles of a reactant such as oxygen is the first step in solving the problem:
Moles of O₂ are calculated using the volume and standard molar volume (22.4 L/mol at STP).
Given the moles of one reactant, stoichiometry helps us find the moles of others involved.
This, combined with the molar mass, allows conversion of moles into grams, as demonstrated to find the mass of aluminum needed.

Gas Laws at STP

Understanding gas laws is key when dealing with gases in chemical reactions.
At Standard Temperature and Pressure (STP), conditions are set to 0°C (273.15 Kelvin) and 1 atm pressure.
These standard conditions simplify calculations by providing a fixed volume of 22.4 L for one mole of gas.

• For the oxygen gas in this reaction, 2.00 L at STP is used to find the moles of O₂.

Using the formula:\[\text{Moles of } O_2 = \frac{\text{Volume of } O_2}{22.4 \text{ L/mol}}\]we find that 2.00 L of oxygen translates to 0.0893 moles.
This concept is critical to converting volumes of gases into moles, facilitating further stoichiometric calculations in chemical equations.