Chemistry for Engineering Students
Chemistry for Engineering Students
4th Edition
ISBN: 9781337398909
Author: Lawrence S. Brown, Tom Holme
Publisher: Cengage Learning
Question
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Chapter 12, Problem 12.116PAE

(a)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the reaction of sodium sulfite and hydrochloric acid should be written.

Concept introduction:

A balanced chemical equation is defined as reaction in which there are same number of constituent atoms on both side of the reaction arrow, that is reactant and product side.

(a)

Expert Solution
Check Mark

Answer to Problem 12.116PAE

Solution: Balanced chemical equation.

NaHSO3(s)+HCl(g)NaCl(s)+SO2(g)+H2O(l)

Explanation of Solution

Here, sulfur dioxide, sodium chloride and water are formed when sodium hydrogen sulfite is reacted with hydrochloric acid. The balanced reaction is as follows:NaHSO3(s)+HCl(g)NaCl(s)+SO2(g)+H2O(l)

(b)

Interpretation Introduction

Interpretation: Mass of SO2 produced if 1.9g of sodium hydrogen sulfite is reacted with excess HCl should be identified.

Concept introduction:

For a reaction, if any reactant is present in excess, the mass of product obtained depends on the limiting reactant. The ratio of number of moles remains the same for any reaction.

Number of mole is related to mass and molar mass as follows:n=mM

where, n is the number of moles

m is mass

M is molar mass

(b)

Expert Solution
Check Mark

Answer to Problem 12.116PAE

Solution: Amount of sulfur dioxide produced is 1.2g

Explanation of Solution

The balanced chemical reaction of sodium hydrogen sulfite and hydrochloride acid is as follows:NaHSO3(s)+HCl(g)NaCl(s)+SO2(g)+H2O(l)

Here, HCl is present in excess, thus, NaHSO3 is a limiting reactant. 1 mole of NaHSO3 produce 1 mole of sulfur dioxide SO2.

Calculate the number of moles of NaHSO3 obtained from 1.9 g of NaHSO3 as follows:n=mM

Molecular mass of NaHSO3=104g/mol

Thus,n NaHSO3=1.9 g104 g/mol=0.0183mol

Since, 1 mole of NaHSO3 produces 1 mole of SO2 thus, 0.0183 mole of NaHSO3 produces  0.0183 mole of SO2

Now, calculate mass of SO2 produced as follows:m=n×M

Molar mass of SO2=64 g/mol

Thus,m SO2=0.0183 mol×64g/mol=1.2g

Therefore, 1.2 g of SO2 is produced from 1.9 g of NaHSO3.

(c)

Interpretation Introduction

Interpretation: Equilibrium concentration of each species in the given reaction should be identified.SO2+12O2SO3K=2.6×1012

Concept introduction: For a general reaction, aA+bBcC+dD

Expression for the equilibrium constant (K) is as follows:K=[C]c[D]d[A]a[B]b

Also, concentration or molarity of a solution is defined as number of moles in 1 L of solution. The unit of concentration or molarity is mol/L or M.

(c)

Expert Solution
Check Mark

Answer to Problem 12.116PAE

Solution: Equilibrium concentration of SO2[SO2]=0M

Equilibrium concentration of SO3[SO3]=0.183M

Equilibrium concentration of O2[O2]=0.41M

Explanation of Solution

Given reaction is SO2+12O2SO3

Equilibrium constant for the reaction will be:k=[SO3][SO2][ O 2]12

The ICE table for the reaction will be:

SO2

12O2

SO3

Initial

0.0183

0.05

0

Change

x

x2

x

equilibrium

0.0183x

0.05x2

x

Since, the value of equilibrium constant is very high thus, SO2 completely react with oxygen to form SO3. The number of moles of SO2 remain will be zero.

Thus,0.0183x=0x=0.0183

Calculate number of moles of O2 remain as follows:

Moles of O2 remain,nO2=(0.05 0.01832)=0.041mole

And, nSO3=0.0183mole

Concentration can be calculated as follows:C=nV

Since, volume of container is 100 mL or 0.1 L, equilibrium concentration of all species can be calculated.[SO2]=n SO 2 V=00.1L=0[O2]=0.0410.1=0.41M[SO3]=0.01830.1=0.183M

Therefore, equilibrium concentration of SO2, O2 and SO3 is 0 M, 0.41 M and 0.183 M respectively.

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Chapter 12 Solutions

Chemistry for Engineering Students

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