×
Log in to StudySoup
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 96ae
Join StudySoup for FREE
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 96ae

Already have an account? Login here
×
Reset your password

Solved: The commercial production of nitric acid involves

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 96AE Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

  • Textbook Solutions
  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

4 5 1 339 Reviews
30
1
Problem 96AE

The commercial production of nitric acid involves the following chemical reactions: (a) Which of these reactions are redox reactions?(b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction, (c) How many grams of ammonia must you start with to make 1000.0 L of a 0.150 M aqueous solution of nitric acid? Assume all the reactions give 100% yield.

Step-by-Step Solution:
Step 1 of 3

Solution: Step1: Redox reactions are the reactions which involves both oxidation and reduction. Oxidation refers to loss of electrons by an atom, molecule or ion whereas reduction means gain of electrons by an atom, molecule or ion. Oxidation leads to increase in oxidation state whereas reduction leads to decrease in oxidation state of the atom, molecule or ion. The reactant which loses electron is said to be oxidized and the reactant which gains electron is said to be reduced. Step2: a) All the reactions involved in the commercial production of nitric acid are redox reaction as they involve both oxidation and reduction. This statement is illustrated in the succeeding steps. Step3: b) In the first reaction, 4NH + 5O ------> 4NO + 6H O, nitrogen undergoes oxidation and 3 2 2 oxygen undergoes reduction. As we have seen, the oxidation number of nitrogen changes from -3 to +2(loss of electrons) i.e., its oxidation number increases. Also, the oxidation number of oxygen decreases from 0 to -2(gain of electrons). Hence, nitrogen is oxidized and oxygen is reduced. Step4: In the second reaction, 2NO + O ----2> 2NO , nitro2n undergoes oxidation and oxygen undergoes reduction. As we have seen, the oxidation number of nitrogen changes from +2 to +4(loss of electrons) i.e., its oxidation number increases. Also, the oxidation number of oxygen decreases from 0 to -4(gain of electrons). Hence, nitrogen is oxidized and oxygen is reduced. Step5: In the third reaction, 3NO + 2 O -2---> 2HNO + NO, 3trogen undergoes oxidation and reduction both. As we have seen, the oxidation number of nitrogen changes from +4 in nitrogen dioxide to +5 in nitric acid(loss of electrons) i.e., its oxidation number increases. Also, the oxidation number of nitrogen decreases from +4 in nitrogen dioxide to +2 in nitrogen monoxide(gain of electrons). Hence, nitrogen is undergoing both oxidation and reduction. Step6: c) Given molarity of nitric acid solution = 0.150 M Volume of nitric acid solution = 1000.0 L Therefore, number of moles of nitric acid = Volume of solution in litres X Molarity = 1000 L X 0.150 M = 150 mol. Now, from the third reaction given above, it is clear that 2 moles of nitric acid is prepared by 3 moles of nitrogen dioxide. Therefore, 150 mol of nitric acid will be prepared by [(3/2) x 150 = 225] mol of nitrogen dioxide. Again, from the second reaction, it is clear that 2 mol of nitrogen dioxide is prepared from 2 mol of nitrogen monoxide. Therefore, 225 mol of nitrogen dioxide will be prepared from 225 mol of nitrogen monoxide. Then, from the first reaction, it is seen that 4 mol of nitrogen monoxide is prepared from 4 mole of ammonia. Therefore, 225 mol of nitrogen monoxide will be prepared from 225 mol of ammonia. Thus number of moles of ammonia required = 225 mol Step7: Now, 1 mol ammonia(NH ) = molar mass of ammonia 3 Therefore, 225 mol ammonia = 225 x molar mass of ammonia = 225 x 17 g [molar mass of ammonia = 17 g/mol] = 3825 g Thus, the required amount of ammonia is 3825 g. ------------------

Step 2 of 3

Chapter 4, Problem 96AE is Solved
Step 3 of 3

Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

Since the solution to 96AE from 4 chapter was answered, more than 1423 students have viewed the full step-by-step answer. This full solution covers the following key subjects: reactions, acid, redox, undergoing, nitric. This expansive textbook survival guide covers 82 chapters, and 9454 solutions. Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247. The answer to “The commercial production of nitric acid involves the following chemical reactions: (a) Which of these reactions are redox reactions?(b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction, (c) How many grams of ammonia must you start with to make 1000.0 L of a 0.150 M aqueous solution of nitric acid? Assume all the reactions give 100% yield.” is broken down into a number of easy to follow steps, and 63 words. The full step-by-step solution to problem: 96AE from chapter: 4 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3.

Other solutions

People also purchased

Related chapters

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

Solved: The commercial production of nitric acid involves