The question is incomplete, here is the complete question:
Consider the following two-step process (unbalanced). How many moles of water can be formed from 0.680 moles of [tex]N_2[/tex] ?
[tex]N_2+H_2\rightarrow NH3[/tex]
[tex]NH_3+O_2\rightarrow NO+H_2O[/tex]
Enter your answer numerically to three siginificant figures.
Answer: The amount of water formed from given amount of nitrogen gas is 1.38 moles
Explanation:
We are given:
Moles of nitrogen gas = 0.46 moles
The balanced chemical equation follows:
[tex]N_2+3H_2\rightarrow 2NH3[/tex] .......(1)
[tex]4NH_3+5O_2\rightarrow 4NO+6H_2O[/tex] .......(2)
From equation 1:
By Stoichiometry of the reaction:
1 mole of nitrogen gas produces 2 moles of ammonia
So, 0.46 moles of nitrogen gas will produce = [tex]\frac{2}{1}\times 0.46=0.92mol[/tex] of ammonia
From equation 2:
By Stoichiometry of the reaction:
4 moles of ammonia produces 6 moles of water
So, 0.92 moles of ammonia will produce = [tex]\frac{6}{4}\times 0.92=1.38mol[/tex] of water
Hence, the amount of water formed from given amount of nitrogen gas is 1.38 moles