Student Solutions Manual For Chemistry: Atoms First
Student Solutions Manual For Chemistry: Atoms First
3rd Edition
ISBN: 9781259923098
Author: Julia Burdge, Jason Overby Professor
Publisher: McGraw-Hill Education
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Chapter 25, Problem 25.43QP
Interpretation Introduction

Interpretation:

Balanced equation for the reaction of phosphorus pentoxide with nitric acid and theoretical yield of dinitrogen pentoxide has to be calculated.

Concept introduction:

Balanced equation

The overall mass of starting materials should be equal to the overall mass of the products.  This is governed by the law of conservation of mass.

Theoretical yield

The yield obtained on a paper by the calculation of reactant amounts taken for that balanced chemical reaction is known as theoretical yield.

Expert Solution & Answer
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Answer to Problem 25.43QP

  • Balanced equation for the reaction of phosphorus pentoxide with nitric acid

4HNO3(aq)+P4O10(s)2N2O5(s)+4HPO3(l)

  • The theoretical yield of dinitrogen pentoxide produced from 79.4g of phosphorus pentoxide with nitric acid is 60.41gofN2O5

Explanation of Solution

To give: The balanced equation for the reaction of phosphorus pentoxide with nitric acid.

Reaction of phosphorus pentoxide with nitric acid

4HNO3(aq)+P4O10(s)2N2O5(s)+4HPO3(l)

Dinitrogen pentoxide is produced from the reaction of phosphorus pentoxide with nitric acid.  The overall mass of the reactants is equal to the overall mass the products.

To calculate: The theoretical yield of dinitrogen pentoxide produced from 79.4g of phosphorus pentoxide with nitric acid.

The amount of phosphorus pentoxide reacted with nitric acid is 79.4g .

Theoreticalyield=reactantamount(g)×(1molofreatantmolarmassofreatant)×(moleratioproductreactant)×(molarmassofproduct1moleofproduct)

Molar mass of phosphorus pentoxide (P4O10) is 283.9g

Molar mass of dinitrogen pentoxide (N2O5) is 108g

Substitute these values into the theoretical yield calculation formula

Theoretical yield of N2O5=79.4g(1moleofP4O10283.9g)×(2moleofN2O51moleofP4O10)×(108gofN2O51moleofN2O5)=(79.4×(2×108283.9))grams=(79.4×0.7608)grams=60.41grams

The theoretical yield of dinitrogen pentoxide produced from 79.4g of phosphorus pentoxide with nitric acid is 60.41gofN2O5

By using molar mass of the reactants and products the theoretical yield of dinitrogen pentoxide produced from the reaction of 79.4g of phosphorus pentoxide with nitric acid was found as 60.41gofN2O5

Conclusion

Balanced equation for the reaction of phosphorus pentoxide with nitric acid and theoretical yield of dinitrogen pentoxide was calculated.

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Chapter 25 Solutions

Student Solutions Manual For Chemistry: Atoms First

Ch. 25 - Prob. 25.11QPCh. 25 - Prob. 25.12QPCh. 25 - Elements 17 and 20 form compounds with hydrogen....Ch. 25 - Prob. 25.14QPCh. 25 - Prob. 25.15QPCh. 25 - Prob. 25.16QPCh. 25 - Prob. 25.17QPCh. 25 - Prob. 25.18QPCh. 25 - Prob. 25.19QPCh. 25 - Prob. 25.20QPCh. 25 - Prob. 25.21QPCh. 25 - Prob. 25.22QPCh. 25 - Prob. 25.23QPCh. 25 - Prob. 25.24QPCh. 25 - Prob. 25.25QPCh. 25 - Prob. 25.26QPCh. 25 - Prob. 25.27QPCh. 25 - Prob. 25.28QPCh. 25 - Prob. 25.29QPCh. 25 - Prob. 25.30QPCh. 25 - Prob. 25.31QPCh. 25 - Prob. 25.32QPCh. 25 - Prob. 25.33QPCh. 25 - Prob. 25.34QPCh. 25 - Prob. 25.35QPCh. 25 - Prob. 25.36QPCh. 25 - Prob. 25.37QPCh. 25 - Prob. 25.38QPCh. 25 - Prob. 25.39QPCh. 25 - Prob. 25.40QPCh. 25 - Prob. 25.41QPCh. 25 - At 620 K, the vapor density of ammonium chloride...Ch. 25 - Prob. 25.43QPCh. 25 - Prob. 25.44QPCh. 25 - Prob. 25.45QPCh. 25 - Prob. 25.46QPCh. 25 - Prob. 25.47QPCh. 25 - Prob. 25.48QPCh. 25 - Prob. 25.49QPCh. 25 - Prob. 25.50QPCh. 25 - Prob. 25.51QPCh. 25 - Prob. 25.52QPCh. 25 - Prob. 25.53QPCh. 25 - Prob. 25.54QPCh. 25 - Prob. 25.55QPCh. 25 - Prob. 25.56QPCh. 25 - Prob. 25.57QPCh. 25 - Prob. 25.58QPCh. 25 - Prob. 25.59QPCh. 25 - Prob. 25.60QPCh. 25 - Prob. 25.61QPCh. 25 - Prob. 25.62QPCh. 25 - Prob. 25.63QPCh. 25 - Prob. 25.64QPCh. 25 - Prob. 25.65QPCh. 25 - Prob. 25.66QPCh. 25 - Prob. 25.67QPCh. 25 - Prob. 25.68QPCh. 25 - Prob. 25.69QPCh. 25 - Prob. 25.70QPCh. 25 - Prob. 25.71QPCh. 25 - Prob. 25.72QPCh. 25 - What are the oxidation numbers of O and F in HFO?Ch. 25 - Prob. 25.74QPCh. 25 - Prob. 25.75QPCh. 25 - Prob. 25.76QPCh. 25 - Prob. 25.77QPCh. 25 - Prob. 25.78QPCh. 25 - Prob. 25.79QPCh. 25 - Prob. 25.80QPCh. 25 - Prob. 25.81QPCh. 25 - Prob. 25.82QPCh. 25 - Prob. 25.83QPCh. 25 - Prob. 25.84QPCh. 25 - Iodine pentoxide (I2O5) is sometimes used to...Ch. 25 - Prob. 25.86QPCh. 25 - Prob. 25.87QPCh. 25 - Prob. 25.88QPCh. 25 - Prob. 25.89QPCh. 25 - Prob. 25.90QPCh. 25 - Prob. 25.91QPCh. 25 - Prob. 25.92QPCh. 25 - Prob. 25.93QPCh. 25 - Prob. 25.94QPCh. 25 - Prob. 25.95QPCh. 25 - Prob. 25.96QPCh. 25 - Prob. 25.97QPCh. 25 - Prob. 25.98QPCh. 25 - Prob. 25.99QPCh. 25 - Prob. 25.100QPCh. 25 - Prob. 25.101QPCh. 25 - Prob. 25.102QPCh. 25 - Prob. 25.103QPCh. 25 - Prob. 25.104QP
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