Answer:
2.7 mol
Explanation:
we have the following reaction
[tex]3 H_2 + N_2\longrightarrow 2 NH_3[/tex]
Data
[tex]2.0 mol H_2\\4.0mol N_2\\mol NH_3= ?[/tex]
First we must calculate what the limit reagent is.
the limit reagent is consumed first, when it is consumed completely the reaction is finished
[tex]3molH_2\longrightarrow 1mol N_2\\4molH_2\longrightarrow x \\x=\frac{4molH_2.1molN_2}{3molH_2}=1.33mol N_2[/tex]
This means that in order to fully consume 4 moles of hydrogen, 1.3 moles of nitrogen are needed, since we have more than this amount of nitrogen (the nitrogen is in excess) is our excess reagent and hydrogen is our limit reagent. so much is it that determines how much product is formed
From the stoichiometric coefficients of the reaction we know that three moles of hydrogen are needed to produce two moles of ammonia
So how many moles of ammonia will be produced if we have 4 moles of hydrogen
To solve it we apply a simple rule of three
[tex]3mol H_2\longrightarrow 2 mol NH_3\\4molH_2\longrightarrow x \\x=\frac{2molNH_3.4molH_2}{3molH_2}=2.7 mol NH_3[/tex]
2.7 mol NH_3 are produced
