   # Metallic molybdenum can be produced from the mineral molybdenite, MoS 2 . 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 × 10 3 kg pure molybdenum from MoS 2 . Assume air contains 21% oxygen by volume, and assume 100% yield for each reaction. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 8, Problem 135AE
Textbook Problem
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## 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.

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

### 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

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