   Chapter 21, Problem 71PS

Chapter
Section
Textbook Problem

In the “contact process” for making sulfuric acid, sulfur is first burned to SO2. Environmental restrictions allow no more than 0.30% of this SO2 to be vented to the atmosphere.(a) If enough sulfur is burned in a plant to produce 1.80 × 106 kg of pure, anhydrous H2SO4 per day, what is the maximum amount of SO2 that is allowed to be exhausted to the atmosphere?(b) One way to prevent any SO2 from reaching the atmosphere is to “scrub” the exhaust gases with slaked lime, Ca(OH)2: Ca(OH) 2 (s) + SO 2 (g)  →  CaSO 3 (s) + H 2 O( l )         2 CaSO 3 (s) + O 2 (g)  →  2 CaSO 4 (s) What mass of Ca(OH)2 (in kilograms) is needed to remove the SO2 calculated in part (a)?

(a)

Interpretation Introduction

Interpretation:

The maximum amount of SO2 that is allowed to be exhausted to the atmosphere is to be calculated.

Concept introduction:

The contact process is the method of producing sulfuric acid in the high concentrations needed for industrial process. The process involved combining of sulfur and oxygen to form sulfur dioxide as follows,

S(s)+O2(g)SO2(g)

Adding an excess of oxygen to sulfur dioxide in the presence of catalyst, the sulfur trioxide is formed. The sulfur trioxide is added to sulfuric acid which gives rise to disulfuric acid (oleum). Oleum is reacted with water to form concentrated sulfuric acid. The reactions involved is written as follows:

2SO2(g)+O2(g)2SO3(g)SO3(g)+H2SO4(l)H2S2O7(l)H2S2O7(l)+H2O(l)2H2SO4(l)                                                             (1)

Numberofmoles=MassMolecularmassMass=Numberofmoles×Molecularmass

Explanation

The maximum amount of SO2 that is allowed to be exhausted to the atmosphere is calculated below.

Given:

The mass of anhydrous sulfuric acid produced per day is 1.8×106kg.

The produced SO2 is only 0.30% allowed to be vented to the atmosphere.

The molecular weight of sulfuric acid is 98gmol1.

The molecular weight of sulfur dioxide is 64gmol1.

The number of moles (n) of sulfuric acid produced is equal to the mass of sulfuric acid produced divided by its molecular weight,

n=1.8×109g98gmol1=0.018×109mol

From equation (1), two moles of concentrated sulfuric acid is produced from two moles of sulfur dioxide. Hence, 0.018×109mol of pure sulfuric acid is produced by 0

(b)

Interpretation Introduction

Interpretation:

To calculate the mass of calcium hydroxide is needed to remove the sulfur dioxide.

Introduction:

In the contact process, for making sulfuric acid, if enough sulfur is burned to sulfur dioxide to produce anhydrous sulfuric acid. The sulfur dioxide is vented to the atmosphere. To prevent any SO2 from reaching atmosphere is to exhaust the gas with slaked lime (Ca(OH)2):

Ca(OH)2(s)+SO2(g)CaSO3(s)+H2O(l)2CaSO3(s)+O2(g)2CaSO4(s) (1)

Numberofmoles=MassMolecularmassMass=Numberofmoles×Molecularmass

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

A(n) always has the same composition.

Introductory Chemistry: A Foundation

Vitamin A supplements can help treat acne. T F

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

Explain what the term nonsuperimposable means.

General, Organic, and Biological Chemistry 