Answer:
[tex]n_{H_2O}=3.0molH_2O\\\\n_{N_2}=1.0molN_2[/tex]
Explanation:
Hello!
In this case, for the described reaction we have:
[tex]2NH_3(g)+\frac{3}{2} O_2(g)\rightarrow N_2(g)+3H_2O(g)[/tex]
Which means there is 2:3/2 mole ratio between ammonia and oxygen and we use it to compute the consumed moles of ammonia by 13.0 moles of oxygen as shown below:
[tex]n_{NH_3}^{consumed \ by\ O_2}=13.0molO_2*\frac{2molNH_3}{\frac{3}{2}molO_2 } =17.33molNH_3[/tex]
However, since just 2.0 mol of ammonia is available, we infer it is the limiting reactant and the maximum amount of both nitrogen and water that can be produced is computed below:
[tex]n_{H_2O}=2.0molNH_3*\frac{3molH_2O}{2molNH_3} =3.0molH_2O\\\\n_{N_2}=2.0molNH_3*\frac{1molN_2}{2molNH_3} =1.0molN_2[/tex]
Best regards!
