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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:

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For each possible product, determine x in equation 1 (Equation 1 and oxidation states for each possible product found in attached images)
N+4-2=+1 2) Ondation state of N in NHs OH": -1 Onidahion stale of N in: N20 N2 NO HNO2 NO2 HNO, +2/2 -2 -2 H +3 -2(2):-4 -2(2):4 H +5 -2(3):-6 +2 +4 +2 +3 +5 ht4
Transcribed Image Text:N+4-2=+1 2) Ondation state of N in NHs OH": -1 Onidahion stale of N in: N20 N2 NO HNO2 NO2 HNO, +2/2 -2 -2 H +3 -2(2):-4 -2(2):4 H +5 -2(3):-6 +2 +4 +2 +3 +5 ht4
reduc of charge from 3+ to 2+) In the case of NH3OH* 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 NH3OH 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: > Fe2+) NH3OH* + H2O → NO2 + 6 H* + 5 e- 5 Fe3+ + NH3OH+ + H2O → 5 Fe2+ + NO2 + 6 H* 5 (Fe3+ + e-→ 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: 2 I 6 Q + Page |3D Tatio, 1.1 1in this case: and so, overall, we would have a umerent stoicfometi 20
Transcribed Image Text:reduc of charge from 3+ to 2+) In the case of NH3OH* 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 NH3OH 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: > Fe2+) NH3OH* + H2O → NO2 + 6 H* + 5 e- 5 Fe3+ + NH3OH+ + H2O → 5 Fe2+ + NO2 + 6 H* 5 (Fe3+ + e-→ 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: 2 I 6 Q + Page |3D Tatio, 1.1 1in this case: and so, overall, we would have a umerent stoicfometi 20
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