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Chapter 5: Problem 63

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

Short Answer

Expert verified
The mass of aluminum reacted is 3.21 g.

Step by step solution

01

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 O鈧� = \( \frac{2.00\text{ L}}{22.4 \text{ L/mol}} \) = 0.0893 mol
02

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 O鈧�. 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 O鈧� calculated in the previous step and solve for the moles of Al: Moles of Al= \( \frac{4}{3} \times 0.0893 \text{ mol} = 0.119 \text{ mol} \)
03

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 脳 molar mass of Al Mass of Al = 0.119 mol 脳 26.98 g/mol = 3.21 g The mass of aluminum reacted is 3.21 g.

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Key Concepts

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

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掳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.

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Most popular questions from this chapter

Urea \(\left(\mathrm{H}_{2} \mathrm{NCONH}_{2}\right)\) is used extensively as a nitrogen source in fertilizers. It is produced commercially from the reaction of ammonia and carbon dioxide: Ammonia gas at \(223^{\circ} \mathrm{C}\) and 90 . atm flows into a reactor at a rate of \(500 . \mathrm{L} / \mathrm{min}\). Carbon dioxide at \(223^{\circ} \mathrm{C}\) and 45 atm flows into the reactor at a rate of \(600 . \mathrm{L} / \mathrm{min}\). What mass of urea is produced per minute by this reaction assuming \(100 \%\) yield?

Atmospheric scientists often use mixing ratios to express the concentrations of trace compounds in air. Mixing ratios are often expressed as ppmv (parts per million volume): ppmv of \(X=\frac{\text { vol of } X \text { at STP }}{\text { total vol of air at STP }} \times 10^{6}\) On a certain November day the concentration of carbon monoxide in the air in downtown Denver, Colorado, reached \(3.0 \times 10^{2}\) ppmv. The atmospheric pressure at that time was 628 torr and the temperature was \(0^{\circ} \mathrm{C}\). a. What was the partial pressure of \(\mathrm{CO} ?\) b. What was the concentration of \(\mathrm{CO}\) in molecules per cubic meter? c. What was the concentration of \(\mathrm{CO}\) in molecules per cubic centimeter?

A sealed balloon is filled with \(1.00 \mathrm{~L}\) helium at \(23^{\circ} \mathrm{C}\) and \(1.00\) atm. The balloon rises to a point in the atmosphere where the pressure is 220 . torr and the temperature is \(-31^{\circ} \mathrm{C}\). What is the change in volume of the balloon as it ascends from \(1.00\) atm to a pressure of 220 . torr?

Hydrogen cyanide is prepared commercially by the reaction of methane, \(\mathrm{CH}_{4}(g)\), ammonia, \(\mathrm{NH}_{3}\left(\mathrm{~g}\right.\) ), and oxygen, \(\mathrm{O}_{2}(\mathrm{~g})\), at high temperature. The other product is gaseous water. a. Write a chemical equation for the reaction. b. What volume of \(\mathrm{HCN}(\mathrm{g})\) can be obtained from the reaction of \(20.0 \mathrm{~L} \mathrm{CH}_{4}(g), 20.0 \mathrm{~L} \mathrm{NH}_{3}(g)\), and \(20.0 \mathrm{~L} \mathrm{O}_{2}(g) ?\) The volumes of all gases are measured at the same temperature and pressure.

The nitrogen content of organic compounds can be determined by the Dumas method. The compound in question is first reacted by passage over hot \(\mathrm{CuO}(s)\) : $$ \text { Compound } \underset{\text { Cvoss }}{\stackrel{\text { Hot }}{\longrightarrow}} \mathrm{N}_{2}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ The product gas is then passed through a concentrated solution of \(\mathrm{KOH}\) to remove the \(\mathrm{CO}_{2}\). After passage through the \(\mathrm{KOH}\) solution, the gas contains \(\mathrm{N}_{2}\) and is saturated with water vapor. In a given experiment a \(0.253-g\) sample of a compound produced \(31.8 \mathrm{~mL} \mathrm{~N}_{2}\) saturated with water vapor at \(25^{\circ} \mathrm{C}\) and 726 torr. What is the mass percent of nitrogen in the compound? (The vapor pressure of water at \(25^{\circ} \mathrm{C}\) is \(23.8\) torr.)
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