Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 8: Problem 65

Consider the following reaction: $$4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)$$ It takes \(2.00 \mathrm{L}\) 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 reacted can be calculated using the following steps: 1. Find the number of moles of oxygen gas: \( \frac{2.00 \, \text{L}}{22.4 \, \text{L/mol}}\) 2. Find the number of moles of aluminum reacted: \( (\text{Moles of O}_2) \times \frac{4 \, \text{moles Al}}{3 \, \text{moles O}_2}\) 3. Calculate the mass of aluminum reacted: \( (\text{Moles of Al}) \times (26.98 \, \text{g/mol})\)

Step by step solution

01

Find the number of moles of oxygen gas

We are given that the volume of oxygen gas at STP is 2.00 L. To find the number of moles, we will use the ideal gas law, which at STP can be simplified to: n = PV/RT At STP, P = 1 atm, V = 2.00 L, R = 0.08206 L atm/mol K, and T = 273.15 K. However, we can simplify the problem since we know 1 mole of any gas occupies 22.4 L at STP. Therefore: Number of moles of oxygen gas = (2.00 L) / (22.4 L/mol)
02

Find the number of moles of aluminum reacted

Now that we have the number of moles of oxygen gas, we can use the stoichiometry of the balanced chemical equation to find the number of moles of aluminum reacted. From the balanced equation, we see that 4 moles of Al reacts with 3 moles of O₂. Hence, we can set up a proportion to determine the moles of aluminum: Moles of Al = (Moles of O₂) × (4 moles Al / 3 moles O₂)
03

Calculate the mass of aluminum reacted

Now that we have the number of moles of aluminum reacted, we can calculate the mass using the molar mass of aluminum, which is approximately 26.98 g/mol. Mass of Al = (Moles of Al) × (26.98 g/mol)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental principle used to relate the pressure, volume, and temperature of an ideal gas. The law is expressed in the equation \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin.
At Standard Temperature and Pressure (STP), which is 1 atm and 273.15 K, the equation can be simplified to find that 1 mole of an ideal gas occupies 22.4 L.
This simplification is useful for quickly finding the number of moles when dealing with gases at STP without needing detailed calculations. To solve problems involving gases at STP:
  • Use 22.4 L/mol directly to find moles from volume.
  • This avoids complex calculation using the full Ideal Gas Law formula.
Balanced Chemical Equation
A balanced chemical equation ensures that the number of atoms for each element is the same on both sides of the equation. This equality maintains the Law of Conservation of Mass. In the equation \( 4 \mathrm{Al}(s) + 3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) \), the coefficients in front of each compound or element show the molar ratio.
This balanced equation indicates:
  • 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide.
  • Understanding these mole ratios is crucial in stoichiometry because they dictate how much of each reactant is needed or how much product is formed.
The balanced equation helps calculate the exact amount of a substance needed or produced in a chemical reaction.
Molar Mass
Molar mass is the mass of one mole of a substance, typically measured in grams per mole (g/mol). For example, the molar mass of aluminum is approximately 26.98 g/mol.
This value is used to convert moles to grams, which is necessary for determining the mass of a substance when the number of moles is known. Steps for calculation:
  • Determine the number of moles using stoichiometry from the balanced equation.
  • Multiply the moles by the molar mass to find the mass in grams.
Molar mass allows you to translate between the count of atoms/molecules and their mass in lab measurements.
Standard Temperature and Pressure (STP)
Standard Temperature and Pressure (STP) are the conditions where the temperature is 273.15 K (0°C) and the pressure is 1 atm. These conditions are commonly used as a reference point in chemistry. At STP:
  • One mole of any ideal gas occupies a volume of 22.4 liters.
  • These standard conditions make it easier to compare gas measurements and perform calculations using the Ideal Gas Law.
STP is particularly helpful when conducting experiments or solving problems involving gases, as it provides a consistent basis for measurement and calculation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The total mass that can be lifted by a balloon is given by the difference between the mass of air displaced by the balloon and the mass of the gas inside the balloon. Consider a hot-air balloon that approximates a sphere \(5.00 \mathrm{m}\) in diameter and contains air heated to \(65^{\circ} \mathrm{C}\). The surrounding air temperature is \(21^{\circ} \mathrm{C}\). The pressure in the balloon is equal to the atmospheric pressure, which is \(745\) torr. a. What total mass can the balloon lift? Assume that the average molar mass of air is \(29.0 \mathrm{g} / \mathrm{mol}\). (Hint: Heated air is less dense than cool air. b. If the balloon is filled with enough helium at \(21^{\circ} \mathrm{C}\) and \(745\) torr to achieve the same volume as in part a, what total mass can the balloon lift? c. What mass could the hot-air balloon in part a lift if it were on the ground in Denver, Colorado, where a typical atmospheric pressure is \(630 .\) torr?

A spherical glass container of unknown volume contains helium gas at \(25^{\circ} \mathrm{C}\) and \(1.960\) atm. When a portion of the helium is withdrawn and adjusted to 1.00 atm at \(25^{\circ} \mathrm{C},\) it is found to have a volume of \(1.75 \mathrm{cm}^{3} .\) The gas remaining in the first container shows a pressure of \(1.710 \) atm. Calculate the volume of the spherical container.

An organic compound contains \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and \(\mathrm{O} .\) Combustion of \(0.1023 \mathrm{g}\) of the compound in excess oxygen yielded \(0.2766 \mathrm{g} \mathrm{CO}_{2}\) and \(0.0991 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) A sample of \(0.4831 \mathrm{g}\) of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129 ). At \(\mathrm{STP}, 27.6 \mathrm{mL}\) of dry \(\mathrm{N}_{2}\) was obtained. In a third experiment, the density of the compound as a gas was found to be \(4.02 \mathrm{g} / \mathrm{L}\) at \(127^{\circ} \mathrm{C}\) and \(256\) torr. What are the empirical and molecular formulas of the compound?

An organic compound containing only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\) yields the following data. i. Complete combustion of \(35.0 \mathrm{mg}\) of the compound produced \(33.5 \mathrm{mg} \mathrm{CO}_{2}\) and \(41.1 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\) ii. A 65.2 -mg sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129 ), giving \(35.6 \mathrm{mL}\) of dry \(\mathrm{N}_{2}\) at \(740 .\) torr and \(25^{\circ} \mathrm{C}\) iii. The effusion rate of the compound as a gas was measured and found to be \(24.6 \mathrm{mUmin.}\) The effusion rate of argon gas, under identical conditions, is \(26.4 \mathrm{mL} / \mathrm{min.}\) What is the molecular formula of the compound?

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of \(200 .\) L/min at \(1.50\) atm and ambient temperature. Air is added to the chamber at 1.00 atm and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g),\) three times as much oxygen as is necessary is reacted. Assuming air is \(21\) mole percent \(\mathrm{O}_{2}\) and \(79\) mole percent \(\mathrm{N}_{2}\), calculate the flow rate of air necessary to deliver the required amount of oxygen. b. Under the conditions in part a, combustion of methane was not complete as a mixture of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) was produced. It was determined that \(95.0 \%\) of the carbon in the exhaust gas was present in \(\mathrm{CO}_{2}\). The remainder was present as carbon in \(\mathrm{CO}\). Calculate the composition of the exhaust gas in terms of mole fraction of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Assume \(\mathrm{CH}_{4}\) is completely reacted and \(\mathrm{N}_{2}\) is unreacted.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free