/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 Autumotive air bags inflate when... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 63

Autumotive air bags inflate when sudium azide, NaN \(_{3}\), rapidly decomposes to its component elements: $$ 2 \mathrm{NaN}_{3}(\mathrm{~s}) \longrightarrow 2 \mathrm{Na}(\mathrm{s})+3 \mathrm{~N}_{2}(g) $$ (a) How many moles of \(\mathrm{N}_{2}\) are produced by the decomposition of \(1.50 \mathrm{~mol}\) of \(\mathrm{NaN}_{3}\) ? (b) How many grams of \(\mathrm{NaN}_{3}\) are required to form \(10.0 \mathrm{~g}\) of nitrogen gas? (c) How many grams of \(\mathrm{NaN}_{3}\) are required to produce \(10.0 \mathrm{ft}^{3}\) of nitrogen gas, about the size of an automotive air bag, if the gas has a density of \(1.25 \mathrm{~g} / \mathrm{L} ?\)

Short Answer

Expert verified
(a) 2.25 moles of N2 are produced when 1.50 moles of NaN3 decompose. (b) 15.5 grams of NaN3 are required to form 10.0 g of N2. (c) 547.6 grams of NaN3 are required to produce 10.0 ft鲁 of N2 with a given density of 1.25 g/L.

Step by step solution

01

Find moles of N2 produced by decomposition of 1.50 mol NaN3

Using the stoichiometry from the balanced chemical equation, we see that for every 2 moles of NaN3, 3 moles of N2 are produced. Therefore, we can use the following proportion to find the moles of N2 produced by the decomposition of 1.50 mol of NaN3: \[ \frac{2 \, \text{moles NaN}_3}{3 \, \text{moles N}_2} = \frac{1.50 \, \text{moles NaN}_3}{x \, \text{moles N}_2} \] Solve for x: \[ x = \frac{3 \times 1.50}{2} = 2.25 \, \text{moles N}_2 \] So, 2.25 moles of N2 are produced when 1.50 moles of NaN3 decompose.
02

Find grams of NaN3 required to form 10.0 g of N2

First, find the number of moles of N2 in 10.0 g of N2: \[ \text{moles of N}_2 = \frac{\text{grams of N}_2}{\text{molar mass of N}_2} \] Molar mass of N2 = \(2 \times 14.01 = 28.02 \,\text{grams per mole}\) \[ \text{moles of N}_2 = \frac{10.0 \, \text{grams}}{28.02 \, \text{grams per mole}} = 0.3568 \, \text{moles N}_2 \] Now, using the stoichiometry from the balanced chemical equation again, we can find the required moles of NaN3: \[ \frac{2 \, \text{moles NaN}_3}{3 \, \text{moles N}_2} = \frac{x \, \text{moles NaN}_3}{0.3568 \, \text{moles N}_2} \] Solve for x: \[ x = \frac{2 \times 0.3568}{3} = 0.2384 \, \text{moles NaN}_3 \] Now, find the mass of NaN3: \[ \text{grams of NaN}_3 = \text{moles of NaN}_3 \times \text{molar mass of NaN}_3 \] Molar mass of NaN3 = \(22.99+3 \times 14.01 = 65.02 \,\text{grams per mole}\) \[ \text{grams of NaN}_3 = 0.2384 \, \text{moles NaN}_3 \times 65.02 \, \text{grams per mole} = 15.5 \,\text{grams NaN}_3 \] So, 15.5 grams of NaN3 are required to form 10.0 g of N2.
03

Find grams of NaN3 required to produce 10.0 ft鲁 of N2 with a given density of 1.25 g/L

First, convert the volume from ft鲁 to L: \[ 10.0 \,\text{ft}^3 \times \frac{28.32 \,\text{L}}{1 \,\text{ft}^3} = 283.2 \,\text{L} \] Find the mass of N2 produced: \[ \text{grams of N}_2 = \text{volume of N}_2 \times \text{density of N}_2 \] \[ \text{grams of N}_2 = 283.2 \, \text{L} \times 1.25 \, \frac{\text{g}}{\text{L}} = 354 \, \text{grams} \] Find the moles of N2: \[ \text{moles of N}_2 = \frac{\text{grams of N}_2}{\text{molar mass of N}_2} = \frac{354 \, \text{grams}}{28.02 \, \text{grams per mole}} = 12.64 \, \text{moles N}_2 \] Using the stoichiometry, find the moles of NaN3: \[ \frac{2 \, \text{moles NaN}_3}{3 \, \text{moles N}_2} = \frac{x \, \text{moles NaN}_3}{12.64 \, \text{moles N}_2} \] Solve for x: \[ x = \frac{2 \times 12.64}{3} = 8.4267 \, \text{moles NaN}_3 \] Now, find the mass of NaN3: \[ \text{grams of NaN}_3 = 8.4267 \, \text{moles NaN}_3 \times 65.02 \, \text{grams per mole} = 547.6 \,\text{grams NaN}_3 \] So, 547.6 grams of NaN3 are required to produce 10.0 ft鲁 of N2 with a given density of 1.25 g/L.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

The Mole-Mole Relationship in Chemical Reactions
Understanding the mole-mole relationship is essential in stoichiometry, the branch of chemistry that involves calculating quantities in chemical reactions. It allows us to predict how much reactant is needed to produce a given amount of product, or vice versa. For instance, when sodium azide decomposes, we can predict the amount of nitrogen gas produced using the balanced chemical equation.

The equation clearly shows a ratio: 2 moles of solid sodium azide yield 3 moles of nitrogen gas. This is a 2:3 mole ratio and is key to solving stoichiometric problems. In the given exercise, from 1.50 moles of NaN3, according to the balanced equation, we can derive 2.25 moles of N2 using the proportion method, highlighting the direct mole-mole relationship that guides the reaction.
Balancing Chemical Equations and Reactions
Balanced chemical equations are the foundation for stoichiometry. They reflect the conservation of mass and the fixed proportions in which chemicals react. For balanced reactions, the number of atoms of each element on the reactant side must equal the number on the product side. In our example with sodium azide decomposing, the balanced equation is written as:
\[2 \mathrm{NaN}_{3}(\mathrm{s}) \longrightarrow 2 \mathrm{Na}(\mathrm{s})+3 \mathrm{N}_{2}(g)\]

This equation indicates that 2 molecules of solid sodium azide decompose to form 2 atoms of solid sodium and 3 molecules of gaseous nitrogen. When solving problems, always ensure you're working with a balanced equation, as this facilitates correct stoichiometric conversions.
Calculating Molar Mass
Molar mass is the mass of 1 mole of a substance and serves as a bridge between the macroscopic world and the atomic scale. It is calculated by summing the atomic masses of all atoms in a formula. For compounds like sodium azide (NaN3), the molar mass is found by adding the atomic masses of one sodium (Na) and three nitrogen (N) atoms together:
\[\text{Molar mass of } \mathrm{NaN}_{3} = 22.99 \text{ (for Na)} + 3 \times 14.01 \text{ (for each N)} = 65.02 \text{ g/mol}\]

With this information, you can convert between moles and grams and vice versa using the formula:
\[\text{mass (g)} = \text{number of moles} \times \text{molar mass (g/mol)}\]
As seen in the exercise, once the molar mass of the substances involved is known, the rest of the stoichiometric calculations follow smoothly.

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

Serotonin is a compound that conducts nerve impulses in the brain. It contains \(68.2\) mass percent \(\mathrm{C}, 6.86\) mass percent \(\mathrm{H}, 15.9\) mass percent \(\mathrm{N}\), and \(9.08\) mass percent O. Its molar mass is \(176 \mathrm{~g} / \mathrm{mol}\). Determine its molecular formula.

The source of oxygen that drives the internal combustion engine in an automobile is air. Air is a mixture of gases, which are principally \(\mathrm{N}_{2}(\sim 79 \%)\) and \(\mathrm{O}_{2}(\sim 20 \%)\). In the cylinder of an automobile engine, nitrogen can react with oxygen to produce nitric oxide gas, NO. As NO is emitted from the tailpipe of the car, it can react with more oxygen to produce nitrogen dioxide gas. (a) Write balanced chemical equations for both reactions. (b) Both nitric oxide and nitrogen dioxide are pollutants that can lead to acid rain and global warming; collectively, they are called "NO \(_{x}\) " gases. In 2004, the United States emitted an estimated 19 million tons of nitrogen dioxide into the atmosphere. How many grams of nitrogen dioxide is this? (c) The production of \(\mathrm{NO}_{\mathrm{x}}\) gases is an unwanted side reaction of the main engine combustion process that turns octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), into \(\mathrm{CO}_{2}\) and water. If \(85 \%\) of the oxygen in an engine is used to combust octane, and the remainder used to produce nitrogen dioxide, calculate how many grams of nitrogen dioxide would be produced during the combustion of 500 grams of octane.

Write the balanced chemical equatiuns fur (a) the complete combustion of acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\), the main active ingredient in vinegar; (b) the decomposition of solid calcium hydroxide into solid calcium(II) oxide (lime) and water vapor; (c) the combination reaction between nickel metal and chlorine gas.

The complete combustion of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), the main component of gasoline, proceeds as follows: $$ 2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \longrightarrow 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) How many moles of \(\mathrm{O}_{2}\) are needed to burn \(1.25 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18} ?\) (b) How many grams of \(\mathrm{O}_{2}\) are needed to burn \(10.0 \mathrm{~g}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}\) ? (c) Octane has a density of \(0.692 \mathrm{~g} / \mathrm{mL}\) at \(20^{\circ} \mathrm{C}\). How many grams of \(\mathrm{O}_{2}\) are required to burn \(1.00\) gal of \(\mathrm{C}_{8} \mathrm{H}_{18}\) ?

The koala dines exclusively on eucalyptus leaves. Its digestive system detoxifies the eucalyptus oil, a poison to other animals. The chief constituent in eucalyptus oil is a substance called eucalyptol, which contains \(77.87 \% \mathrm{C}\), \(11.76 \% \mathrm{H}\), and the remainder \(\mathrm{O}\). (a) What is the empirical formula for this substance? (b) A mass spectrum of eucalyptol shows a peak at about 154 amu. What is the molecular formula of the substance?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Nuclear 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.