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In the important industrial process for producing ammonia (the Haber Process), the overall reaction is:

[tex]\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) + 100.4 \, \text{kJ} \][/tex]

A yield of \( NH_3 \) of approximately \( 98\% \) can be obtained at \( 200^{\circ}C \) and 1,000 atmospheres of pressure. How many grams of \( N_2 \) must react to form 1.7 grams of ammonia, \( NH_3 \)?

A. \( 0.052 \, \text{g} \)
B. \( 1.4 \, \text{g} \)
C. \( 0.0058 \, \text{g} \)
D. [tex]\( 28 \, \text{g} \)[/tex]



Answer :

GinnyAnswer
To find out how many grams of \( N_2 \) are required to produce 1.7 grams of ammonia (\( NH_3 \)), we can follow a systematic approach using stoichiometry and the molar masses of the substances involved.

1. Determine the Molar Masses:
- Molar mass of \( N_2 \): 28.02 g/mol
- Molar mass of \( NH_3 \): 17.03 g/mol

2. Calculate the Moles of \( NH_3 \) Produced:
- Desired mass of \( NH_3 \): 1.7 grams
- Moles of \( NH_3 \) can be calculated using its molar mass:
[tex]\[ \text{Moles of } NH_3 = \frac{\text{mass of } NH_3}{\text{molar mass of } NH_3} = \frac{1.7 \text{ grams}}{17.03 \text{ g/mol}} \][/tex]
This calculation gives approximately 0.09982 moles of \( NH_3 \).

3. Use the Stoichiometry of the Balanced Reaction:
- According to the balanced chemical equation:
[tex]\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \][/tex]
- This reaction tells us that 1 mole of \( N_2 \) produces 2 moles of \( NH_3 \).

4. Calculate the Moles of \( N_2 \) Needed:
- From the stoichiometry of the reaction:
[tex]\[ \text{Moles of } N_2 = \frac{\text{Moles of } NH_3}{2} = \frac{0.09982 \text{ moles}}{2} \][/tex]
This calculation gives approximately 0.04991 moles of \( N_2 \).

5. Convert Moles of \( N_2 \) to Grams:
- Using the molar mass of \( N_2 \):
[tex]\[ \text{Mass of } N_2 = \text{moles of } N_2 \times \text{molar mass of } N_2 = 0.04991 \text{ moles} \times 28.02 \text{ g/mol} \][/tex]
This calculation gives approximately 1.3985 grams of \( N_2 \).

Therefore, to form 1.7 grams of ammonia ([tex]\( NH_3 \)[/tex]), you need approximately 1.4 grams of nitrogen gas ([tex]\( N_2 \)[/tex]). Thus, the closest correct choice is [tex]\(\boxed{1.4 \text{ g}}\)[/tex].

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