Ammonia and oxygen react to form nitrogen and water, like this: 4NH3(g)+ 3O2(g)=

Q: Ammonia and oxygen react to form nitrogen and water, like this: 4NH3(g)+ 3O2(g)=

1) 2N2(g) + 6H2O(g) 4NH3(g) + 3O2(g) 2) 0.22 mol NH3 and 0.165 mol O2 are there 1 mol NH3 needs 0.75 mol O2 again reactants are stoichiometric as 1 mol NH3 give 0.5 mol N2 so 0.11 mol N2 will be there is flask 3) 0.24 mol N2 and 0.72 mol O2 1 mol N2 needs 3 mol O2, so reactants are stoichiometric as per stoichiometry no N2 will be present in flask, it will get converted to 0.48 mol NH3

ID: 723694 • Letter: A

Question

Ammonia and oxygen react to form nitrogen and water, like this:

4NH3(g)+ 3O2(g)= 2N2(g) 6H2O(g)
Use this chemical equation to answer the questions in the table below.NH

1) write a balanced equation, including physical state for the reverse reaction.

2) suppose 220. mmol of NH3 and165. mmol of O2 are added to an empty flask how much
N2 would be in the flask at equilibrium?

3) suppose 240. mmol of N2 and 720. mmol of H2O are added to an empty flask how much
N2 would be in the flask at equilibrium?

Explanation / Answer

1)
2N2(g) + 6H2O(g) <=> 4NH3(g) + 3O2(g)

0.22 mol NH3 and 0.165 mol O2 are there

1 mol NH3 needs 0.75 mol O2

again reactants are stoichiometric

as 1 mol NH3 give 0.5 mol N2

so 0.11 mol N2 will be there is flask

0.24 mol N2 and 0.72 mol O2

1 mol N2 needs 3 mol O2,

so reactants are stoichiometric

as per stoichiometry no N2 will be present in flask, it will get converted to 0.48 mol NH3

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Ammonia and oxygen react to form nitrogen and water, like this: 4NH3(g)+ 3O2(g)=

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