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Chapter 5: Problem 71

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

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

Step by step solution

01

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 O鈧� = Volume of O鈧� / Volume occupied by 1 mole of O鈧� at STP Moles of O鈧� = 2.00 L / 22.4 L/mol Moles of O鈧� = 0.0893 mol
02

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 O鈧� 鈫� 2 Al鈧侽鈧� From the balanced equation, we can see that 3 moles of O鈧� react with 4 moles of Al. Using this ratio, we can calculate the number of moles of aluminum: Moles of Al = (moles of O鈧� 脳 moles of Al) / moles of O鈧� Moles of Al = (0.0893 mol 脳 4 mol) / 3 mol Moles of Al = 0.1191 mol
03

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 脳 molar mass of Al Mass of Al = 0.1191 mol 脳 26.98 g/mol Mass of Al = 3.214 g Therefore, the mass of aluminum that reacted with oxygen gas is 3.214 g.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 O鈧�(g) 鈫� 2 Al鈧侽鈧�(s)
This equation tells us that aluminum (Al) reacts with oxygen (O鈧�) to form aluminum oxide (Al鈧侽鈧�).
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 脳 10虏鲁).
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 O鈧� 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掳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 O鈧�.
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.

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Most popular questions from this chapter

A mixture of cyclopropane and oxygen is sometimes used as a general anesthetic. Consider a balloon with an anesthetic mixture of cyclopropane and oxygen at 170. torr and 570. torr, respectively. Calculate the mole fraction of cyclopropane in the mixture.

An aerosol can contains 400. mL of compressed gas at 5.20 atm. When all of the gas is sprayed into a large plastic bag, the bag inflates to a volume of 2.14 L. What is the pressure of gas in the plastic bag? Assume a constant temperature.

Consider two different containers, each filled with 2 moles of \(\mathrm{Ne}(g)\) . One of the containers is rigid and has constant volume. The other container is flexible (like a balloon) and is capable of changing its volume to keep the external pressure and internal pressure equal to each other. If you raise the temperature in both containers, what happens to the pressure and density of the gas inside each container? Assume a constant external pressure.

Consider separate \(2.5-\) L gaseous samples of \(\mathrm{He}\), \(\mathrm{N}_{2},\) and \(\mathrm{F}_{2},\) all at \(\mathrm{STP}\) and all acting ideally. Rank the gases in order of increasing average kinetic energy and in order of increasing average velocity.

A person accidentally swallows a drop of liquid oxygen, \(\mathrm{O}_{2}(l)\) which has a density of 1.149 \(\mathrm{g} / \mathrm{mL}\) . Assuming the drop has a volume of 0.050 \(\mathrm{mL}\) , what volume of gas will be produced in the person's stomach at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and a pressure of 1.0 \(\mathrm{atm} ?\)
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