Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Textbook Question
Chapter 8, Problem 135AE

Metallic molybdenum can be produced from the mineral molybdenite, MoS2. The mineral is first oxidized in air to molybdenum trioxide and sulfur dioxide. Molybdenum trioxide is then reduced to metallic molybdenum using hydrogen gas. The balanced equations are

MoS 2 ( s )   + 7 2 O 2 ( g ) MoO 3 ( s ) +  2SO 2 ( g ) MoO 3 ( s )   +  3H 2 ( g )  Mo ( s )   +  3H 2 O ( l )

Calculate the volumes of air and hydrogen gas at 17°C and 1.00 atm that are necessary to produce 1.00 × 103 kg pure molybdenum from MoS2. Assume air contains 21% oxygen by volume, and assume 100% yield for each reaction.

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

The volume of air and volume of H2 required to producing 1×103Kg of molybdenum from MoS2 in the given reactions at 17oC and 1atm are to be calculated.

Concept introduction:

  • Number of moles of a substance from its given mass is,

    Number of moles=GivenmassMolecularmass

  • Mole ratios between the reactants of a reaction are depends upon the coefficients of respective reactant in a balanced chemical equation.
  • According to ideal gas equation for finding volume of a gas,

Volume=Numberofmoles×R×Temperaturepressure

Answer to Problem 135AE

Answer

The volume of air required to producing 1×103Kg of molybdenum is 4.1×106L.

The volume of H2 required to producing 1×103Kg of molybdenum is 7.42×105L.

Explanation of Solution

Explanation

  • To find: the number of moles of producing 1×103Kg of molybdenum.

The number of moles of molybdenum produced is 1.04×104mol.

The given produced mass of molybdenum is 1×103Kg=1×106g

So the number of moles of molybdenum produced is,

The number of moles of a substance from its given mass is,

Number of moles=GivenmassMolecularmass

Therefore, the number of moles of molybdenum produced is,

Number of molesof sulfur=1×106g×1mole95.94g=1.04×104mol

Hence, the number of moles of molybdenum produced is 1.04×104mol.

  • To find: the molar ratio between molybdenum to H2, MoO3 and O2 in the given reactions.

Molar ratio between molybdenum to H2 is 1:3.

Molar ratio between molybdenum to MoO3 is 1:1.

Molar ratio between molybdenum to O2 is 1:72.

The balance equation of given reactions are,

MoS2+72O2MoO3+2SO2

MoO3+3H2Mo+3H2O

Here, in second reaction,

1 mole of MoO3 is reacted with 3 mole of H2 to produce 1 mole of molybdenum in the reaction.

Therefore,

The molar ratio between molybdenum to MoO3 is 1:1.

The molar ratio between molybdenum to H2 is 1:3.

In first reaction,

1 mole of MoS2 is reacted with 72mol O2 to produce 1 mole of MoO3 in the reaction.

Therefore,

The mole ratio between MoO3 to O2 is 1:7.

Since, the molar ratio between molybdenum to MoO3 is 1:1.

Hence, the molar ratio between molybdenum to O2 is 1:72.

  • To find: the number of moles of H2, O2 in the given reactions.

The number of moles of H2 is 3.12×104mol.

The number of moles of O2 is 3.64×104mol

The molar ratio between molybdenum to H2 is calculated as 1:3.

The molar ratio between molybdenum to O2 is calculated as 1:72.

The number of moles of molybdenum is 1.04×104mol.

Therefore,

  • The number of moles of H2 is,

1.04×104molMo×3molH21molMo=3.12×104molH2

  • The number of moles of O2 is,

1.04×104molMo×72molO21molMo=3.36×104molO2

Hence,

The number of moles of H2 in the given reactions is 3.12×104mol.

The number of moles of O2 in the given reactions is 3.64×104mol

  • To determine: the volume of H2 required to producing 1×103Kg of molybdenum from MoS2 in the given reactions

The volume of H2 required to producing 1×103Kg of molybdenum is 7.42×105L.

The number of number of moles of H2 reacted in the given reactions is 3.12×104mol.

The pressure is given as 1atm.

The temperature is given as 17oC=290K.

The equation for finding volume of a gas from its number of moles is,

According to ideal gas equation for finding volume of a gas,

Volume=Numberofmoles×R×Temperaturepressure

Therefore,

Volumeof H2=3.12×104mol×0.08206Latm/Kmol×290K1atm =7.42×105L

Hence, the volume of H2 required to producing 1×103Kg of molybdenum is 7.42×105L.

  • To determine: volume of air required to producing 1×103Kg of molybdenum from MoS2 in the given reactions

The volume of air required to producing 1×103Kg of molybdenum is 4.1×106L.

The number of moles of O2 in the given reactions is calculated as 3.64×104mol

The pressure is given as 1atm.

The temperature is given as 17oC=290K.

The equation for finding volume of a gas from its number of moles is,

According to ideal gas equation for finding volume of a gas,

Volume=Numberofmoles×R×Temperaturepressure

Volumeof O2=3.64×104mol×0.08206Latm/Kmol×290K1atm =8.66×105L

      

The 21% volume of air containing O2 by volume.

That is,

volumeofair×21100=volumeofO2

Therefore,

volumeofair=100×8.66×105L21 =4.1×106L

Hence, the volume of air required to producing 1×103Kg of molybdenum is 4.1×106L.

Conclusion

Conclusion

The volume of air and volume of H2 required to producing 1×103Kg of molybdenum from MoS2 in the given reactions are determined by ideal gas equation. The number of moles of H2 and O2 are calculated by checking their mole ratio between produced molybdenum.

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Chapter 8 Solutions

Chemistry: An Atoms First Approach

Ch. 8 - Prob. 3ALQCh. 8 - Prob. 4ALQCh. 8 - Prob. 6ALQCh. 8 - Prob. 8ALQCh. 8 - Prob. 11ALQCh. 8 - Prob. 12ALQCh. 8 - Prob. 15ALQCh. 8 - Prob. 16ALQCh. 8 - Draw molecular-level views that show the...Ch. 8 - Prob. 20QCh. 8 - Prob. 21QCh. 8 - Prob. 22QCh. 8 - Prob. 23QCh. 8 - Prob. 24QCh. 8 - Prob. 25QCh. 8 - Consider two different containers, each filled...Ch. 8 - Prob. 27QCh. 8 - Prob. 28QCh. 8 - Prob. 29QCh. 8 - Prob. 30QCh. 8 - Prob. 31QCh. 8 - Prob. 32QCh. 8 - Prob. 33QCh. 8 - Prob. 34QCh. 8 - Prob. 35QCh. 8 - Prob. 36QCh. 8 - Prob. 37ECh. 8 - Prob. 38ECh. 8 - A sealed-tube manometer (as shown below) can be...Ch. 8 - Prob. 40ECh. 8 - A diagram for an open-tube manometer is shown...Ch. 8 - Prob. 42ECh. 8 - Prob. 43ECh. 8 - Prob. 44ECh. 8 - Prob. 45ECh. 8 - Prob. 46ECh. 8 - Prob. 47ECh. 8 - Prob. 48ECh. 8 - Prob. 49ECh. 8 - Prob. 50ECh. 8 - The Steel reaction vessel of a bomb calorimeter,...Ch. 8 - A 5.0-L flask contains 0.60 g O2 at a temperature...Ch. 8 - Prob. 53ECh. 8 - A person accidentally swallows a drop of liquid...Ch. 8 - A gas sample containing 1.50 moles at 25C exerts a...Ch. 8 - Prob. 56ECh. 8 - Prob. 57ECh. 8 - What will be the effect on the volume of an ideal...Ch. 8 - Prob. 59ECh. 8 - Prob. 60ECh. 8 - An ideal gas is contained in a cylinder with a...Ch. 8 - Prob. 62ECh. 8 - A sealed balloon is filled with 1.00 L helium at...Ch. 8 - Prob. 64ECh. 8 - Consider the following reaction:...Ch. 8 - A student adds 4.00 g of dry ice (solid CO2) to an...Ch. 8 - Air bags are activated when a severe impact causes...Ch. 8 - Concentrated hydrogen peroxide solutions are...Ch. 8 - In 1897 the Swedish explorer Andre tried to reach...Ch. 8 - Sulfur trioxide, SO3, is produced in enormous...Ch. 8 - A 15.0-L rigid container was charged with 0.500...Ch. 8 - An important process for the production of...Ch. 8 - Consider the reaction between 50.0 mL liquid...Ch. 8 - Urea (H2NCONH2) is used extensively as a nitrogen...Ch. 8 - Prob. 75ECh. 8 - Prob. 76ECh. 8 - Prob. 77ECh. 8 - A compound has the empirical formula CHCl. 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You...Ch. 8 - Prob. 160CPCh. 8 - You are given an unknown gaseous binary compound...Ch. 8 - Prob. 162CPCh. 8 - Calculate w and E when 1 mole of a liquid is...Ch. 8 - The preparation of NO2(g) from N2(g) and O2(g) is...Ch. 8 - In the presence of nitric acid, UO2+ undergoes a...Ch. 8 - Silane, SiH4, is the silicon analogue of methane,...Ch. 8 - Prob. 167IPCh. 8 - Prob. 168IPCh. 8 - Prob. 169MP
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