Respuesta :
Answer : The [tex]N_2[/tex] element is the reactant in excess.
Solution : Given,
Moles of [tex]N_2[/tex] = 5 moles
Moles of [tex]H_2[/tex] = 3.50 moles
The balanced chemical reaction is,
[tex]3H_2(g)+N_2(g)\rightarrow 2NH_3(g)[/tex]
From the balanced reaction we conclude that
As, 3 moles of [tex]H_2[/tex] react with 1 mole of [tex]N_2[/tex]
So, 3.5 moles of [tex]H_2[/tex] react with [tex]\frac{3.5}{3}=1.16[/tex] moles of [tex]N_2[/tex]
The excess of [tex]N_2[/tex] = 5 - 1.16 = 3.84 moles
That means in the given balanced reaction, [tex]H_2[/tex] is a limiting reagent because it limits the formation of products and [tex]N_2[/tex] is an excess reagent.
Hence, the [tex]N_2[/tex] element is the reactant in excess.
Given 3.50 moles of hydrogen and 5.00 moles of nitrogen to produce ammonia, nitrogen is the excess reactant.
What is the excess reactant?
The excess reactant is the reactant in a chemical reaction with a greater amount than necessary to react completely with the limiting reactant.
- Step 1: Write the balanced equation.
3 H₂ + N₂ ⇒ 2 NH₃
- Step 2: Establish the theoretical ratio.
The theoretical ratio (TR) of H₂ to N₂ is 3:1.
- Step 3: Establish the experimental ratio.
The experimental ratio (ER) of H₂ to N₂ is 3.50:5.00 = 0.70:1.
- Step 4: Determine the excess reactant.
Comparing TR and ER, we can realize that there is not enough hydrogen to react with the nitrogen. Thus, nitrogen is the excess reactant.
Given 3.50 moles of hydrogen and 5.00 moles of nitrogen to produce ammonia, nitrogen is the excess reactant.
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