Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 14, Problem 14.114P

(a)

Interpretation Introduction

Interpretation:

The three steps of Ostwald process has to be given.

Concept-Introduction:

Ostwald process: It is a chemical process where ammonium is converted to nitric acid.

Oxidation reaction: The loss of electrons or the gain of oxigen atoms. And also increase their oxidation number.

Disproportionation reaction: It is type of reaction where a substance can act as a reducing agent as well as an oxidizing agent since an atom present in the substance reacts to give atoms having and lower oxidation states.

(a)

Expert Solution
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Explanation of Solution

Oxidation takes place in the first two step of Ostwald process. Ammonium oxidizes to NO which further oxidizes to give NO2.  In the third step, nitric acid is formed by a disproportionation reaction.

The three steps of Ostwald process follows as,

  4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)2NO(g) + O2(g)  2NO2(g)3NO2(g) + H2O(l)  2HNO3(l) + NO(g)

(b)

Interpretation Introduction

Interpretation:

Moles of NH3 are consumed per mole of HNO3 produced has to be determined.

Concept-Introduction:

Ostwald process: It is a chemical process where ammonium is converted to nitric acid.

(b)

Expert Solution
Check Mark

Explanation of Solution

The three steps of Ostwald process follows as,

  4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)2NO(g) + O2(g)  2NO2(g)3NO2(g) + H2O(l)  2HNO3(l) + NO(g)

Moles of NH3 are consumed per mole of HNO3 produced when NO is not recycled is determined as given,

No.of moles of NH3 consumedNo.of moles of HNO3 produced = (3 mol NO22 mol HNO3)(2 mol NO2 mol NO2)(4 mol NH34 mol NO)= 1.5 mol NH3/mol HNO3

Moles of NH3 are consumed per mole of HNO3 produced is 1.5 mol NH3/mol HNO3 .

(c)

Interpretation Introduction

Interpretation:

Volume of concentrated aqueous nitric acid has to be determined using the given data.

Concept-Introduction:

Ostwald process: It is a chemical process where ammonium is converted to nitric acid.

According to Ideal gas equation,

Numberofmoles=Pressure×VolumeR×Temperature

(c)

Expert Solution
Check Mark

Explanation of Solution

Given data is shown below:

  Yield in the first step = 96%Yield in the second and third step = 100%Moles of are consumed per mole of produced= 1.5 mol NH3/mol HNO3Pressure = 5.0 atmVolume = 1 m3Temperature= 850oC = (273+850)K= 1123KDensity = 1.37g/mol

Number of moles of NH3 is calculated as given,

No. of moles of NH3, n = PVRT= (5.0 atm)(1 m3)(0.0821 L.atm.mol1.K1)(1123 K)(1 L103 m3)(10% NH3100% gas)= 5.42309 mol NH3

Number of moles of NH3 is 5.42309 mol NH3.

Mass of HNO3 is determined as follows,

Mass of HNO3 can be obtained from the number of moles of NH3 present where it is converted to moles HNO3.

Mass (g) of HNO3 = (5.42309 mol NH3)(1 mol HNO31.5 mol NH3)(63.02 g HNO31 mol HNO3)(96%100%)= 218.728 g HNO3

Volume of concentrated aqueous nitric acid is calculated as shown below,

Volume of HNO3 = (218.728 g HNO3)(100%60%)(1 mL1.37 g)= 2.7×102 mL HNO3

Volume of concentrated aqueous nitric acid is 2.7×102 mL.

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Chapter 14 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

Ch. 14 - Prob. 14.11PCh. 14 - Prob. 14.12PCh. 14 - Prob. 14.13PCh. 14 - Prob. 14.14PCh. 14 - Prob. 14.15PCh. 14 - Prob. 14.16PCh. 14 - Prob. 14.17PCh. 14 - Prob. 14.18PCh. 14 - Prob. 14.19PCh. 14 - Prob. 14.20PCh. 14 - Prob. 14.21PCh. 14 - Prob. 14.22PCh. 14 - Prob. 14.23PCh. 14 - Prob. 14.24PCh. 14 - Prob. 14.25PCh. 14 - Prob. 14.26PCh. 14 - Prob. 14.27PCh. 14 - Prob. 14.28PCh. 14 - Prob. 14.29PCh. 14 - Prob. 14.30PCh. 14 - Prob. 14.31PCh. 14 - Prob. 14.32PCh. 14 - Prob. 14.33PCh. 14 - Prob. 14.34PCh. 14 - Prob. 14.35PCh. 14 - Prob. 14.36PCh. 14 - Prob. 14.37PCh. 14 - Prob. 14.38PCh. 14 - Prob. 14.39PCh. 14 - Prob. 14.40PCh. 14 - Prob. 14.41PCh. 14 - Prob. 14.42PCh. 14 - Prob. 14.43PCh. 14 - Prob. 14.44PCh. 14 - Prob. 14.45PCh. 14 - Prob. 14.46PCh. 14 - Give explanations for the large drops in melting...Ch. 14 - Prob. 14.48PCh. 14 - Prob. 14.49PCh. 14 - Prob. 14.50PCh. 14 - Prob. 14.51PCh. 14 - Prob. 14.52PCh. 14 - Prob. 14.53PCh. 14 - Prob. 14.54PCh. 14 - Prob. 14.55PCh. 14 - Prob. 14.56PCh. 14 - Prob. 14.57PCh. 14 - Prob. 14.58PCh. 14 - Prob. 14.59PCh. 14 - Prob. 14.60PCh. 14 - Prob. 14.61PCh. 14 - Prob. 14.62PCh. 14 - Prob. 14.63PCh. 14 - Prob. 14.64PCh. 14 - Prob. 14.65PCh. 14 - Prob. 14.66PCh. 14 - Prob. 14.67PCh. 14 - Prob. 14.68PCh. 14 - Prob. 14.69PCh. 14 - Prob. 14.70PCh. 14 - Prob. 14.71PCh. 14 - Prob. 14.72PCh. 14 - Prob. 14.73PCh. 14 - Prob. 14.74PCh. 14 - Prob. 14.75PCh. 14 - Prob. 14.76PCh. 14 - Prob. 14.77PCh. 14 - Prob. 14.78PCh. 14 - Prob. 14.79PCh. 14 - Prob. 14.80PCh. 14 - Prob. 14.81PCh. 14 - Prob. 14.82PCh. 14 - Prob. 14.83PCh. 14 - Prob. 14.84PCh. 14 - Prob. 14.85PCh. 14 - Prob. 14.86PCh. 14 - Prob. 14.87PCh. 14 - Prob. 14.88PCh. 14 - Prob. 14.89PCh. 14 - Prob. 14.90PCh. 14 - Prob. 14.91PCh. 14 - Prob. 14.92PCh. 14 - Prob. 14.93PCh. 14 - Prob. 14.94PCh. 14 - Prob. 14.95PCh. 14 - Prob. 14.96PCh. 14 - Prob. 14.97PCh. 14 - Prob. 14.98PCh. 14 - Prob. 14.99PCh. 14 - Prob. 14.100PCh. 14 - Prob. 14.101PCh. 14 - Prob. 14.102PCh. 14 - Prob. 14.103PCh. 14 - Prob. 14.104PCh. 14 - Xenon tetrafluoride reacts with antimony...Ch. 14 - Prob. 14.106PCh. 14 - Prob. 14.107PCh. 14 - Prob. 14.108PCh. 14 - Prob. 14.109PCh. 14 - Prob. 14.110PCh. 14 - Prob. 14.111PCh. 14 - Prob. 14.112PCh. 14 - Prob. 14.113PCh. 14 - Prob. 14.114PCh. 14 - Prob. 14.115PCh. 14 - Prob. 14.116PCh. 14 - Prob. 14.117PCh. 14 - Prob. 14.118PCh. 14 - Prob. 14.119PCh. 14 - Prob. 14.120PCh. 14 - Prob. 14.121PCh. 14 - Prob. 14.122PCh. 14 - Prob. 14.123PCh. 14 - Prob. 14.124PCh. 14 - Prob. 14.125PCh. 14 - Prob. 14.126PCh. 14 - Prob. 14.127PCh. 14 - Prob. 14.128PCh. 14 - Prob. 14.129PCh. 14 - Prob. 14.130PCh. 14 - Prob. 14.131PCh. 14 - Prob. 14.132PCh. 14 - Prob. 14.133PCh. 14 - Prob. 14.134PCh. 14 - Hydrogen peroxide can act as either an oxidizing...
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