Answer:

Explanation:
N2(g)+O2(g)⇌2NO(g), 
N2(g)+2H2(g)⇌N2H4(g), 
2H2O(g)⇌2H2(g)+O2(g), 
If we add above reaction we will get:
2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)
Equilibrium constant for Eq (1) is 
Divide Eq (1) by 2, it will become:
N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)
Equilibrium constant for Eq (2) is 

Answer:
2.318032g
Explanation:
-The electrolysis equation of water is written as below:

-The mole ratio of Water to the hydrogen formed is 1:1, therefore 2.3 moles of hydrogen gas is produced.
-Hydrogen's molar mass is 1.00784 grams:

Hence, 2.318032 grams of hydrogen is produced.