Question

Consider the following reaction: $$4\mathrm{Al}(s)+3{\mathrm{O}}_2(g)⟶2{\mathrm{Al}}_2{\mathrm{O}}_3(s)$$ It takes $2.00\mathrm{L}$ 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 reacted is 3.21 g.

Step-by-step solution

Convert the volume of oxygen to moles

The volume of pure oxygen gas at STP (standard temperature and pressure) is given as 2.00 L. At STP, 1 mole of any gas occupies a volume of 22.4 L. We can use this information to find out how many moles of oxygen gas are in 2.00 L: $\text{Moles of }{O}_2=\frac{\text{Volume of }{O}_2}{\text{Volume occupied by 1 mole at STP}}$ Moles of O₂ = $\frac{2.00\text{ L}}{22.4\text{ L/mol}}$ = 0.0893 mol

Determine the moles of aluminum required

Next, we need to use the stoichiometry of the balanced equation to determine the moles of aluminum required to react with the given amount of oxygen. From the balanced equation, we can see that 4 moles of Al react with 3 moles of O₂. We can set up the following proportion: $\frac{\text{Moles of Al}}{\text{Moles of }{O}_2}=\frac{4}{3}$ Now, we can plug in the moles of O₂ calculated in the previous step and solve for the moles of Al: Moles of Al= $\frac{4}{3}\times 0.0893\text{ mol}=0.119\text{ mol}$

Convert moles of aluminum to grams

Finally, we need to convert the moles of aluminum to grams to find the mass of aluminum reacted. The molar mass of aluminum is 26.98 g/mol. We can use this molar mass to convert moles of Al to grams: Mass of Al = moles of Al × molar mass of Al Mass of Al = 0.119 mol × 26.98 g/mol = 3.21 g The mass of aluminum reacted is 3.21 g.

Key concepts

Chemical Reaction Stoichiometry

Understanding chemical reaction stoichiometry is pivotal when dealing with reactions, as it ties together the proportion of reactants and products. Stoichiometry is grounded in the law of conservation of mass which states that matter cannot be created or destroyed in an isolated system. The balance in a chemical equation represents this, ensuring that the number of atoms for each element is equal on the reactant and product sides.

For instance, in the reaction provided in the exercise, the coefficients 4, 3, and 2 tell us directly the molar ratio in which aluminum (\text{Al}) and oxygen (\text{O${}_2$}) react to form aluminum oxide (\text{Al${}_2$O${}_3$}). This ratio allows us to solve complex chemistry problems by calculating the amount of reactants needed or products formed.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we observe. One mole of any substance contains Avogadro's number (\text{6.022 x 10${}^{23}$}) of particles, which could be atoms, molecules, ions, or electrons.

In the exercise, moles are utilized to convert the volume of oxygen gas to a quantity that can be used alongside the stoichiometric coefficients of the chemical equation. Understanding this concept allows the student to manipulate quantities of substances in a chemical reaction with precision.

Gas Laws

The gas laws are a series of laws that relate the pressure, volume, temperature, and number of moles of a gas. At standard temperature and pressure (0°C and 1 atm), one mole of any gas occupies 22.4 liters. This is a key point from the ideal gas law, which is a cornerstone of understanding how gases behave under different conditions.

In the context of our exercise, the gas law is applied to determine that 2 liters of oxygen gas at STP corresponds to 0.0893 moles. Knowledge of gas laws is also essential in predicting how gases will react under changing conditions.

Molar Mass

The molar mass is the mass of one mole of a substance, typically measured in grams per mole (g/mol). It is numerically equivalent to the atomic or molecular weight of a substance but gives us a way to translate moles to grams, an invaluable step for practical lab work.

As seen in the final calculation step for the aluminum mass in our exercise, the molar mass of aluminum (26.98 g/mol) is crucial for converting the calculated moles of aluminum into grams, thus providing the mass of aluminum that reacted with the oxygen.