Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 22

Write a balanced equation for the preparation of (a) \(\mathrm{N}_{2}\) from \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2}\) (b) \(\mathrm{O}_{2}\) from \(\mathrm{O}_{3}\) (c) \(\mathrm{S}\) from \(\mathrm{H}_{2} \mathrm{~S}\)

Short Answer

Expert verified
Question: Write balanced chemical equations for the preparation of the following elements: (a) Nitrogen (N2) from Lead Azide (Pb(N3)2) (b) Oxygen (O2) from Ozone (O3) (c) Sulfur (S) from Hydrogen Sulfide (H2S). Answer: (a) 3Pb(N3)2 → 6N2 + 2Pb (b) 2O3 → 3O2 (c) 2H2S → S2 + 2H2

Step by step solution

01

Write unbalanced equation

Write the equation with the reactant and product: Pb(N3)2 -> N2
02

Balance the equation

Balancing the equation requires multiplying the reactants or products with coefficients as needed to ensure an equal number of atoms on both sides of the equation. 3Pb(N3)2 -> 6N2 + 2Pb
03

Finalize the balanced equation

The balanced equation for the preparation of N2 from Pb(N3)2 is: 3Pb(N3)2 → 6N2 + 2Pb (b) Preparation of O2 from O3
04

Write unbalanced equation

Write the equation with the reactant and product: O3 -> O2
05

Balance the equation

Balancing the equation requires multiplying the reactants or products with coefficients as needed to ensure an equal number of atoms on both sides of the equation. 2O3 -> 3O2
06

Finalize the balanced equation

The balanced equation for the preparation of O2 from O3 is: 2O3 → 3O2 (c) Preparation of S from H2S
07

Write unbalanced equation

Write the equation with the reactant and product: H2S -> S
08

Balance the equation

Balancing the equation requires multiplying the reactants or products with coefficients as needed to ensure an equal number of atoms on both sides of the equation. 2H2S -> S2 + 2H2
09

Finalize the balanced equation

The balanced equation for the preparation of S from H2S is: 2H2S → S2 + 2H2

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

Write a balanced net ionic equation for the disproportionation reaction (a) of iodine to give iodate and iodide ions in basic solution. (b) of chlorine gas to chloride and perchlorate ions in basic solution.

Choose the strongest acid from each group. (a) \(\mathrm{HClO}, \mathrm{HBrO}, \mathrm{HIO}\) (b) \(\mathrm{HIO}, \mathrm{HIO}_{3}, \mathrm{HIO}_{4}\) (c) \(\mathrm{HIO}, \mathrm{HBrO}_{2}, \mathrm{HBrO}_{4}\)

Given $$\begin{array}{ll} \mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q) & K_{\mathrm{a}}=6.9 \times 10^{-1} \\ \mathrm{HF}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{HF}_{2}-(a q) & K=2.7 \end{array}$$ calculate \(K\) for the reaction $$ 2 \mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HF}_{2}^{-}(a q) $$

Write a balanced net ionic equation for the reaction of nitric acid with (a) a solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\). (b) \(\mathrm{Ag}(s)\); assume the nitrate ion is reduced to \(\mathrm{NO}_{2}(g)\) (c) \(\mathrm{Cd}(s)\); assume the nitrate ion is reduced to \(\mathrm{N}_{2}(g)\).

Give the Lewis structure of (a) \(\mathrm{NO}_{2}\) (b) \(\mathrm{NO}\) (c) \(\mathrm{SO}_{2}\) (d) \(\mathrm{SO}_{3}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

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