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A reaction vessel for synthesizing ammonia by reacting nitrogen and hydrogen is filled with 6.04 kg H2 and excess N2. If 28.0 kg NH3 is produced, what is the percent yield of the reaction?

Respuesta :

Rohan11

Answer:

81.8 %

Explanation:

The balanced equation for the reaction is [tex]3H_2 + N_2[/tex]  ⇒  [tex]2 NH_3[/tex]

So to find the number of moles of [tex]H_2[/tex] we do 6.04kg ÷ 2 = 3.02 moles.

The mole ratio between [tex]H_2[/tex] and [tex]NH_3[/tex] is 3:2. So to find the number of moles you do  [tex]\frac{3.02}{3}[/tex] × 2 which gives you 2.0124 moles.

Then you find the mass by doing moles multiplied by Mr:  2.0124 x 17 which gives you 34.21 kg.

Then to find the percentage yield you do:   [tex]\frac{28}{34.21}[/tex] x 100 = 81.82.......

Which is 81.8 %

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