In the case of NH30H+ the situation is far more complex. You will consider N2, N20, NO, HNO2, NO2, and HNO3 as possible products. For now, we may describe the reaction as follows: x Fe3+ + NH3OH* x Fe2+ +? (1) x = moles of Fe3+ reacting (or Fe2+ produced) per mole of NH3OH* ? = unknown product (N2, N20, NO, HNO2, NO2, or HNO3) If, for example, the nitrogen-containing product were NO2, the oxidation half-equation for NH30H would be: NH30H + H20 → NO2 + 6 H* + 5 e- In that case, the overall balanced equation for the reaction of NH3OH and Fe3+ would be: 5 (Fe3+ + e- Fe2*) NH3OH* + H2O → NO2 + 6 H* + 5e- 5 Fe3+ + NH30H+ + H2O →5 Fe2+ + NO2 + 6 H* Thus, the stoichiometric ratio of Fe3+ to NH30H in this reaction would be 5:1. On the other hand, if the product were N2, then we would have: | 6 Q + Page 2 and so, overall, we would havea dferemt stoiciometric Tatio, 1.1 in this case:
In the case of NH30H+ the situation is far more complex. You will consider N2, N20, NO, HNO2, NO2, and HNO3 as possible products. For now, we may describe the reaction as follows: x Fe3+ + NH3OH* x Fe2+ +? (1) x = moles of Fe3+ reacting (or Fe2+ produced) per mole of NH3OH* ? = unknown product (N2, N20, NO, HNO2, NO2, or HNO3) If, for example, the nitrogen-containing product were NO2, the oxidation half-equation for NH30H would be: NH30H + H20 → NO2 + 6 H* + 5 e- In that case, the overall balanced equation for the reaction of NH3OH and Fe3+ would be: 5 (Fe3+ + e- Fe2*) NH3OH* + H2O → NO2 + 6 H* + 5e- 5 Fe3+ + NH30H+ + H2O →5 Fe2+ + NO2 + 6 H* Thus, the stoichiometric ratio of Fe3+ to NH30H in this reaction would be 5:1. On the other hand, if the product were N2, then we would have: | 6 Q + Page 2 and so, overall, we would havea dferemt stoiciometric Tatio, 1.1 in this case:
Chapter7: Statistical Data Treatment And Evaluation
Section: Chapter Questions
Problem 7.9QAP
Related questions
Question
100%
For each possible product, determine x in equation 1
(Equation 1 and oxidation states for each possible product found in attached images)
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you