Respuesta :

tprogamer477
To balance any redox equation, first divide the reaction into two half reactions, and balance the atom whose oxidation number is changing: 

NO2 --> NH3 
H2 --> H2O 

Next, balance oxygens by adding H2O where needed, and then balance hydrogen by adding H+: 

NO2 --> NH3 + 2 H2O 
H2O + H2 --> H2O 


7 H+ + NO2 --> NH3 + 2 H2O 
H2O + H2 --> H2O + 2 H+ 

Next, balance charges by adding electrons (e-): 

7 H+ + NO2 + 7 e- --> NH3 + 2 H2O 
H2O + H2 --> H2O + 2 H+ + 2 e- 

Now, multiply one or both equations by small numbers to make the number of electrons the same in the two half reactions. In this case, multiply the top equation by 2 and the bottom one by 7 to give you 14 electrons in each half reaction: 

14 H+ + 2 NO2 + 14 e- --> 2 NH3 + 4 H2O 
7 H2O + 7 H2 --> 7 H2O + 14 H+ + 14 e- 

Finally, add the two half reactions and simplify by cancelling anything that is on both sides. This will give you the final balanced reaction: 

2 NO2 + 7 H2 --> 2 NH3 + 4 H2O 

So, the answer to the original question is (A) 7

Otras preguntas