If 100.0g of nitrogen is reacted with 100.0g of hydrogen, what is the theoretical yield of the reaction in grams? What is the excess reactant? What is the limiting reactant?

66.7 g of NH3, hydrogen is the limiting, nitrogen is in excess.
121.6 g of NH3, nitrogen is the limiting, hydrogen is in excess.
33.0 mol of NH3, nitrogen is the limiting, hydrogen is in excess.
121.6 g of NH3, hydrogen is the limiting, nitrogen is in excess.

Respuesta :

Answer:

121.6 of [tex]NH_{3}[/tex] ,Nitrogen is limiting and Hydrogen is in excess.

Step-by-step explanation:

The reaction involved in this question is known as Haeber's Process and the reaction is as follows

[tex]N_{2} + 3H_{2} -->2 NH_{3}[/tex]

We need to find the moles of each to determine the total product, the excess reagent and limiting reagent.

No. of moles of[tex]N_{2}[/tex]=[tex]\frac{100}{28}[/tex]=3.57 mol

No. of moles of [tex]H_{2}[/tex]=[tex]\frac{100}{2}[/tex]=50 moles

We need moles of [tex]H_{2}[/tex] and [tex]N_{2}[/tex] in the ratio of 3:1.

Therefore, we can clearly see that [tex]H_{2}[/tex]  is excess. And [tex]N_{2}[/tex] is the limitng reagent.

As, [tex]N_{2}[/tex] is the limiting reagent a total of 2 * 3.57 moles of [tex]NH_{3}[/tex] will be formed.

Therefore, total mass of [tex]NH_{3}[/tex] =7.14*17=121.6 g

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