Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN: 9781305079373
Author: William L. Masterton, Cecile N. Hurley
Publisher: Cengage Learning
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Chapter 5, Problem 91QAP
Interpretation Introduction

(a)

Interpretation:

By burning sulfur with oxygen sulfur trioxide can be prepared:

2S(s) + 3O2(g)  2SO3(g)

A 5.00 L flask having 5 g of oxygen and sulfur at a temperature of 25 0C and pressure of 995 mm Hg is heated. The temperature in the flask increases to 138 0C after the completion of reaction. The pressure of SO3 in the flask needs to be calculated.

Concept introduction:

According to the ideal gas law volume i.e. V, pressure i.e. P, number of moles i.e. n, temperature i.e. T and Universal gas constant i.e. R are interrelated as below:

PV = nRT

The molarity of the solution is calculated as follows:

M=nV(L)

Here, n is number of moles of solute and V is volume of solution in L.

Expert Solution
Check Mark

Answer to Problem 91QAP

1.05 atm

Explanation of Solution

By burning sulfur with oxygen sulfur trioxide can be prepared

2S(s) + 3O2(g)  2SO3(g)

In the 5 g of sulfur the mole of sulfur present is:

ns=ms(MM)s    =5 g32.0 g/mol   =0.156 mol

Moles of sulfur reacted is 0.156 mol.

Similarly, number of moles of oxygen can be calculated using the ideal gas equation as follows:

n=PVRT

The given pressure can be converted into atm as follows:

1 mm Hg=0.001316 atm

Thus,

995 mm Hg=1.31 atm

The given temperature can be converted into Kelvin as follows:

0 C=273.15 K

Thus,

25 C=298.15 K

Putting the values,

n=(1.31 atm)(5 L)(0.082 atm L K1 mol1)(298.15 K)=0.268mol

From the balanced chemical reaction, 2 mol of S must react with 3 mol of O2 thus, 0.156 mol of S must react with 32×0.156=0.233 mol.

The number of moles of oxygen gas given is 0.268 mol thus, it is present in excess and the sulfur is limiting reactant.

The number of moles of SO3 produced is same as that of sulfur.

After completion of reaction, the volume is 5.00 L, temperature is 138 0C and the pressure in the flask due to SO3 is calculated as below:

PSO3V=nSO3RTPSO3(5.00 L)=(0.156 mol)(0.08205 L atm K1 mol1)(138 0C)              PSO3= (0.156 mol)(0.08205 L atm K1 mol1)(138+273)K5.00 L                    = 1.05 atm

Therefore, the pressure of SO3 is 1.05 atm.

Interpretation Introduction

(b)

Interpretation:

The total pressure in the flask needs to be determined.

Concept introduction:

The ideal gas equation is as follows:

PV=nRT

Here, P is pressure, V is volume, n is number of moles, R is Universal gas constant and T is temperature.

Expert Solution
Check Mark

Answer to Problem 91QAP

1.27 atm.

Explanation of Solution

The molar ratio of O2 :S is 3:2. The moles of oxygen used to react with 0.156 mol of sulfur are calculated as below:

(nO2)reacted=ns(Molar ratio of O2 and S)                = 0.156 mol(32)                =0.234 mol

Therefore, the moles of O2 used in 0.156 mol of sulfur are 0.234 mol.

The pressure in the flask prior the reaction is 995 mm Hg. The moles of O2 present initially a 25 0C are calculated as below:

PO2V=(nO2)excessRT(995 mm Hg)(5.00 L)=(nO2)excess(0.08205 L atm K1 mol1)(25 0C)                    (nO2)excess= (995 mm Hg)(5.00 L)(0.08205 L atm K1 mol1)(25+273)K(0.0013151 mmHg)                                  = 0.268 mol

Therefore, mole of oxygen present in the mixture is 0.268 mol.

The oxygen which is left unreacted is calculated as below:

(nO2)unreacted=(nO2)excess(nO2)reacted                  = 0.268 mol  0.234 mol                  = 0.034 mol

The total number of moles in the flask =

ntot=nSO3(nO2)reacted      = 0.156 mol + 0.034 mol      = 0.190 mol

The total number of moles in the flask is 0.190 mol.

The total pressure of gas in the flask due to SO3 and unreacted oxygen is calculated as

PtotV=ntotRTPtot(5.00 L)=(0.190 mol)(0.08205 L atm K1 mol1)(138 0C)              Ptot= (0.190 mol)(0.08205 L atm K1 mol1)(138+273)K5.00 L                   = 1.27 atm

Therefore, the total pressure in the flask is 1.27 atm.

Interpretation Introduction

(c)

Interpretation:

The molarity of H2 SO4 formed if 250.0 mL is added into the flask needs to be determined.

Concept introduction:

According to the ideal gas law volume i.e. V, pressure i.e. P, number of moles i.e. n, temperature i.e. T and Universal gas constant i.e. R are interrelated as below:

PV = nRT

The molarity of the solution is calculated as follows:

M=nV(L)

Here, n is number of moles of solute and V is volume of solution in L.

Expert Solution
Check Mark

Answer to Problem 91QAP

0.03 M.

Explanation of Solution

The volume of the flask containing SO3 is 5.00 L, if 250.0 mL of water is added, the total volume of solution becomes:

VSolution=5.00 L+2501000 L=5.25 L

The molarity of solution is calculated as follows:

Molarity of H2SO4 = moles of solute (SO3)volume of solution (H2SO4)

The moles of SO3 is 0.156 mol. The molarity of H2 SO4 is calculated as below:

Molarity of H2SO4 = 0.156 mol5.25 L                              = 0.03 M

Therefore, the molarity of H2 SO4 is 0.03 M.

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Chapter 5 Solutions

Chemistry: Principles and Reactions

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