Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 31

When mixtures of gaseous \(\mathrm{H}_{2}\) and gaseous \(\mathrm{Cl}_{2}\) react, a product forms that has the same properties regardless of the relative amounts of \(\mathrm{H}_{2}\) and \(\mathrm{Cl}_{2}\) used. a. How is this result interpreted in terms of the law of definite proportion? b. When a volume of \(\mathrm{H}_{2}\) reacts with an equal volume of \(\mathrm{Cl}_{2}\) at the same temperature and pressure, what volume of product having the formula \(\mathrm{HCl}\) is formed?

Short Answer

Expert verified
a. The reaction between $\mathrm{H}_{2}$ and $\mathrm{Cl}_{2}$ to form $\mathrm{HCl}$ abides by the law of definite proportion as the product always contains the same elements, H and Cl, in the same proportion, 1:1, regardless of the relative amounts of $\mathrm{H}_{2}$ and $\mathrm{Cl}_{2}$. b. When an equal volume of $\mathrm{H}_{2}$ reacts with $\mathrm{Cl}_{2}$ at the same temperature and pressure, double the volume of product $\mathrm{HCl}$ is formed. If the volume of $\mathrm{H}_{2}$ is V (liters) and the volume of $\mathrm{Cl}_{2}$ is also V (liters), they react to form a volume of $\mathrm{HCl}$ equal to 2V (liters).

Step by step solution

01

Understand the Law of Definite Proportion

The Law of Definite Proportion states that a chemical compound always contains the same elements in the same proportions, regardless of the method of preparation or the source of the material. In this case, we need to explain the given result in terms of the Law of Definite Proportion.
02

Interpretation of the reaction

According to the problem, the product formed (HCl) has the same properties regardless of the relative amounts of H2 and Cl2. This indicates that the product formed always contains the same elements, hydrogen (H) and chlorine (Cl), in the same proportions, 1:1 in this case. Thus, the reaction abides by the law of definite proportion.
03

Reaction between gaseous H2 and Cl2

Let's analyze the balanced chemical equation of the reaction between gaseous hydrogen (H2) and gaseous chlorine (Cl2) to form hydrogen chloride (HCl): \[H_{2(g)} + Cl_{2(g)} \rightarrow 2 HCl_{(g)}\]
04

Volume relationship of gases in the reaction

By reaction stoichiometry, we can determine the mole relationship of H2, Cl2, and HCl. According to the balanced equation, one mole of H2 reacts with one mole of Cl2 to form two moles of HCl. At the same temperature and pressure, one mole of any gas occupies the same volume. Hence, if the volume of H2 is V (liters) and the volume of Cl2 is also V (liters), they react to form double the volume of HCl, which is 2V (liters).
05

Answer the questions

a. The fact that a product with the same properties is formed regardless of the relative amounts of H2 and Cl2 supports the Law of Definite Proportion. It indicates that the product formed (HCl) always contains the same elements, H and Cl, in the same proportion, 1:1. b. When an equal volume of H2 reacts with Cl2 at the same temperature and pressure, double the volume of product HCl is formed. If the volume of H2 is V (liters) and the volume of Cl2 is also V (liters), they react to form a volume of HCl equal to 2V (liters).

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!

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

You have two distinct gaseous compounds made from element \(\mathrm{X}\) and element \(\mathrm{Y}\). The mass percents are as follows: Compound I: \(30.43 \% \mathrm{X}, 69.57 \% \mathrm{Y}\) Compound II: \(63.64 \% \mathrm{X}, 36.36 \% \mathrm{Y}\) In their natural standard states, element \(\mathrm{X}\) and element \(\mathrm{Y}\) exist as gases. (Monatomic? Diatomic? Triatomic? That is for you to determine.) When you react "gas X" with "gas Y" to make the products, you get the following data (all at the same pressure and temperature): 1 volume "gas \(\mathrm{X}^{\prime \prime}+2\) volumes "gas \(\mathrm{Y}^{\prime \prime} \longrightarrow\) 2 volumes compound \(I\) 2 volumes "gas \(\mathrm{X}^{\prime \prime}+1\) volume "gas \(\mathrm{Y}^{\prime \prime} \longrightarrow\) 2 volumes compound II Assume the simplest possible formulas for reactants and products in the chemical equations above. Then, determine the relative atomic masses of element \(\mathrm{X}\) and element \(\mathrm{Y}\).

What is the modern view of the structure of the atom?

What are the symbols of the following metals: sodium, radium, iron, gold, manganese, lead.

What is the systematic name of \(\mathrm{Ta}_{2} \mathrm{O}_{5}\) ? If the charge on the metal remained constant and then sulfur was substituted for oxygen, how would the formula change? What is the difference in the total number of protons between \(\mathrm{Ta}_{2} \mathrm{O}_{5}\) and its sulfur analog?

Write the formula for each of the following compounds: a. chromium(III) hydroxide c. lead(IV) carbonate b. magnesium cyanide d. ammonium acetate
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

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