Answered

Ammonia is produced by the following reaction.
[tex]\[
3 H_2(g) + N_2(g) \rightarrow 2 NH_3(g)
\][/tex]

When 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because:

A. 7.5 mol of hydrogen would be needed to consume the available nitrogen.
B. 7.5 mol of nitrogen would be needed to consume the available hydrogen.
C. Hydrogen would produce 7.5 mol more ammonia than nitrogen.
D. Nitrogen would produce 7.5 mol more ammonia than hydrogen.



Answer :

GinnyAnswer
Certainly! Let's break down the problem step-by-step to determine how much ammonia is produced and to confirm why hydrogen is the limiting reactant.

### Balancing the Reaction

The balanced chemical equation is given as:
[tex]\[ 3H_2(g) + N_2(g) \rightarrow 2NH_3(g) \][/tex]

### Calculating Moles

Firstly, we need to convert the mass of [tex]\( H_2 \)[/tex] and [tex]\( N_2 \)[/tex] to moles.

- Molar masses (g/mol):
- For [tex]\( H_2 \)[/tex]: [tex]\( 2.016 \)[/tex] g/mol
- For [tex]\( N_2 \)[/tex]: [tex]\( 28.014 \)[/tex] g/mol

- Given masses:
- [tex]\( H_2 \)[/tex] mass: [tex]\( 7.00 \)[/tex] g
- [tex]\( N_2 \)[/tex] mass: [tex]\( 70.0 \)[/tex] g

- Moles of [tex]\( H_2 \)[/tex]:
[tex]\[ \text{Moles of } H_2 = \frac{\text{Mass of } H_2}{\text{Molar mass of } H_2} = \frac{7.00 \text{ g}}{2.016 \text{ g/mol}} = 3.472 \text{ moles} \][/tex]

- Moles of [tex]\( N_2 \)[/tex]:
[tex]\[ \text{Moles of } N_2 = \frac{\text{Mass of } N_2}{\text{Molar mass of } N_2} = \frac{70.0 \text{ g}}{28.014 \text{ g/mol}} = 2.498 \text{ moles} \][/tex]

### Stoichiometric Calculations

1. Calculate how much [tex]\( H_2 \)[/tex] is needed to react with the available [tex]\( N_2 \)[/tex]:

The balanced equation states 3 moles of [tex]\( H_2 \)[/tex] react with 1 mole of [tex]\( N_2 \)[/tex].

Required [tex]\( H_2 \)[/tex] for the available [tex]\( N_2 \)[/tex]:
[tex]\[ \text{Needed } H_2 = 2.498 \text{ moles of } N_2 \times \frac{3 \text{ moles of } H_2}{1 \text{ mole of } N_2} = 7.496 \text{ moles of } H_2 \][/tex]

2. Calculate how much [tex]\( N_2 \)[/tex] is needed to react with the available [tex]\( H_2 \)[/tex]:

The balanced equation states 3 moles of [tex]\( H_2 \)[/tex] react with 1 mole of [tex]\( N_2 \)[/tex].

Required [tex]\( N_2 \)[/tex] for the available [tex]\( H_2 \)[/tex]:
[tex]\[ \text{Needed } N_2 = 3.472 \text{ moles of } H_2 \times \frac{1 \text{ mole of } N_2}{3 \text{ moles of } H_2} = 1.157 \text{ moles of } N_2 \][/tex]

### Determining the Limiting Reactant

- Comparing the moles required:
- The available [tex]\( H_2 \)[/tex] is 3.472 moles.
- We need 7.496 moles of [tex]\( H_2 \)[/tex] to react with all of [tex]\( N_2 \)[/tex] (more than available).
- The available [tex]\( N_2 \)[/tex] is 2.498 moles.
- We need 1.157 moles of [tex]\( N_2 \)[/tex] to react with all of [tex]\( H_2 \)[/tex] (less than available).

Clearly, [tex]\( H_2 \)[/tex] is the limiting reactant because we don't have enough [tex]\( H_2 \)[/tex] to react with all the [tex]\( N_2 \)[/tex].

### Ammonia Production

- From the balanced equation, 3 moles of [tex]\( H_2 \)[/tex] produce 2 moles of [tex]\( NH_3 \)[/tex].
- Moles of [tex]\( NH_3 \)[/tex] produced from the limiting reactant ([tex]\( H_2 \)[/tex]):

[tex]\[ \text{Moles of } NH_3 = 3.472 \text{ moles of } H_2 \times \frac{2 \text{ moles of } NH_3}{3 \text{ moles of } H_2} = 2.315 \text{ moles of } NH_3 \][/tex]

- Converting moles of [tex]\( NH_3 \)[/tex] to grams (Molar mass of [tex]\( NH_3 \)[/tex] is [tex]\( 17.031 \)[/tex] g/mol):

[tex]\[ \text{Mass of } NH_3 = 2.315 \text{ moles} \times 17.031 \text{ g/mol} = 39.424 \text{ g} \][/tex]

### Summary
- [tex]\( H_2 \)[/tex] is the limiting reactant because 7.5 moles of [tex]\( H_2 \)[/tex] would be needed to react with the available [tex]\( N_2 \)[/tex] (more than we have).
- Actual ammonia produced: [tex]\( 2.315 \)[/tex] moles or [tex]\( 39.424 \)[/tex] grams.

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