Answer:
[tex]0.30molNH_3[/tex]
Explanation:
Hello there!
In this case, since the reaction for the formation of ammonia is:
[tex]3H_2+N_2\rightarrow 2NH_3[/tex]
We can evidence the 1:2 mole ratio of nitrogen gas to ammonia; therefore, the appropriate stoichiometric setup for the calculation of the moles of the latter turns out to be:
[tex]0.15molN_2*\frac{2molNH_3}{1molN_2}[/tex]
And the result is:
[tex]0.30molNH_3[/tex]
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