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You are carrying out the following reaction:
[tex]\[N_2 + 3 H_2 \longrightarrow 2 NH_3\][/tex]

You start with 6.0 moles of nitrogen gas and 6.0 moles of hydrogen gas. How many moles of ammonia, [tex][tex]$NH_3$[/tex][/tex], will you make?

A. 16.0 moles
B. 6.0 moles
C. 4.0 moles
D. 3.0 moles
E. 5.0 moles

Sagot :

GinnyAnswer
In order to determine the number of moles of ammonia ([tex]\( NH_3 \)[/tex]) produced from the reaction starting with 6.0 moles of nitrogen gas ([tex]\( N_2 \)[/tex]) and 6.0 moles of hydrogen gas ([tex]\( H_2 \)[/tex]), we need to analyze the stoichiometry of the reaction.

The balanced chemical equation for the reaction is:
[tex]\[ N_2 + 3H_2 \rightarrow 2NH_3 \][/tex]

From this equation, the stoichiometric coefficients tell us that:
- 1 mole of [tex]\( N_2 \)[/tex] reacts with 3 moles of [tex]\( H_2 \)[/tex] to produce 2 moles of [tex]\( NH_3 \)[/tex].

First, let’s calculate how many moles of [tex]\( NH_3 \)[/tex] we can get from the given moles of [tex]\( N_2 \)[/tex]:
- According to the equation, 1 mole of [tex]\( N_2 \)[/tex] produces 2 moles of [tex]\( NH_3 \)[/tex].
- Thus, 6.0 moles of [tex]\( N_2 \)[/tex] would ideally produce [tex]\( 6.0 \times 2 = 12.0 \)[/tex] moles of [tex]\( NH_3 \)[/tex].

Next, let’s calculate how many moles of [tex]\( NH_3 \)[/tex] we can get from the given moles of [tex]\( H_2 \)[/tex]:
- According to the equation, 3 moles of [tex]\( H_2 \)[/tex] are needed to produce 2 moles of [tex]\( NH_3 \)[/tex].
- Thus, 6.0 moles of [tex]\( H_2 \)[/tex] would ideally produce [tex]\( (6.0 / 3) \times 2 = 4.0 \)[/tex] moles of [tex]\( NH_3 \)[/tex].

To determine the actual number of moles of [tex]\( NH_3 \)[/tex] produced, we need to identify the limiting reagent. The limiting reagent is the reactant that will be completely consumed first, and thus determines the maximum amount of product that can be formed.

Compare the calculated amounts:
- [tex]\( NH_3 \)[/tex] from [tex]\( N_2 \)[/tex]: 12.0 moles
- [tex]\( NH_3 \)[/tex] from [tex]\( H_2 \)[/tex]: 4.0 moles

Since the [tex]\( H_2 \)[/tex] produces fewer moles of [tex]\( NH_3 \)[/tex], [tex]\( H_2 \)[/tex] is the limiting reagent.

Therefore, the number of moles of [tex]\( NH_3 \)[/tex] produced will be limited by the [tex]\( H_2 \)[/tex] and will be:
[tex]\[ 4.0 \text{ moles} \][/tex]

So, the answer is:
[tex]\[ 4.0 \text{ moles of } NH_3 \][/tex]
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