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Answer :
To determine the theoretical yield of ammonia ([tex]\(NH_3\)[/tex]) formed in the reaction between nitrogen ([tex]\(N_2\)[/tex]) and hydrogen ([tex]\(H_2\)[/tex]), we can follow these steps:
1. Determine the Molar Masses:
- The molar mass of nitrogen ([tex]\(N_2\)[/tex]) is about 28.02 g/mol.
- The molar mass of hydrogen ([tex]\(H_2\)[/tex]) is about 2.02 g/mol.
- The molar mass of ammonia ([tex]\(NH_3\)[/tex]) is about 17.03 g/mol.
2. Calculate the Moles of Each Reactant:
- For [tex]\(N_2\)[/tex]:
[tex]\[
\text{Moles of } N_2 = \frac{30.5 \, \text{g}}{28.02 \, \text{g/mol}} \approx 1.09 \, \text{moles}
\][/tex]
- For [tex]\(H_2\)[/tex]:
[tex]\[
\text{Moles of } H_2 = \frac{8.65 \, \text{g}}{2.02 \, \text{g/mol}} \approx 4.28 \, \text{moles}
\][/tex]
3. Identify the Limiting Reactant:
- According to the balanced equation, [tex]\(1\)[/tex] mole of [tex]\(N_2\)[/tex] reacts with [tex]\(3\)[/tex] moles of [tex]\(H_2\)[/tex].
- Calculate the moles of [tex]\(H_2\)[/tex] needed for the moles of [tex]\(N_2\)[/tex] available:
[tex]\[
\text{Moles of } H_2 \text{ needed} = 1.09 \, \text{moles of } N_2 \times 3 \approx 3.27 \, \text{moles}
\][/tex]
- Compare the available moles of [tex]\(H_2\)[/tex] (4.28 moles) to the moles needed (3.27 moles). Since 4.28 moles is greater than 3.27 moles, [tex]\(N_2\)[/tex] is the limiting reactant.
4. Calculate the Moles of [tex]\(NH_3\)[/tex] Formed:
- Use the limiting reactant [tex]\(N_2\)[/tex] to determine the [tex]\(NH_3\)[/tex] produced.
- According to the stoichiometry: [tex]\(1\)[/tex] mole of [tex]\(N_2\)[/tex] produces [tex]\(2\)[/tex] moles of [tex]\(NH_3\)[/tex].
[tex]\[
\text{Moles of } NH_3 = 1.09 \, \text{moles of } N_2 \times 2 = 2.18 \, \text{moles of } NH_3
\][/tex]
5. Calculate the Theoretical Yield of [tex]\(NH_3\)[/tex] in Grams:
- Finally, convert moles of [tex]\(NH_3\)[/tex] to grams:
[tex]\[
\text{Grams of } NH_3 = 2.18 \, \text{moles of } NH_3 \times 17.03 \, \text{g/mol} \approx 37.07 \, \text{g}
\][/tex]
Thus, the theoretical yield of [tex]\(NH_3\)[/tex] is approximately 37.1 g, matching the option "37.1 g NH_3".
1. Determine the Molar Masses:
- The molar mass of nitrogen ([tex]\(N_2\)[/tex]) is about 28.02 g/mol.
- The molar mass of hydrogen ([tex]\(H_2\)[/tex]) is about 2.02 g/mol.
- The molar mass of ammonia ([tex]\(NH_3\)[/tex]) is about 17.03 g/mol.
2. Calculate the Moles of Each Reactant:
- For [tex]\(N_2\)[/tex]:
[tex]\[
\text{Moles of } N_2 = \frac{30.5 \, \text{g}}{28.02 \, \text{g/mol}} \approx 1.09 \, \text{moles}
\][/tex]
- For [tex]\(H_2\)[/tex]:
[tex]\[
\text{Moles of } H_2 = \frac{8.65 \, \text{g}}{2.02 \, \text{g/mol}} \approx 4.28 \, \text{moles}
\][/tex]
3. Identify the Limiting Reactant:
- According to the balanced equation, [tex]\(1\)[/tex] mole of [tex]\(N_2\)[/tex] reacts with [tex]\(3\)[/tex] moles of [tex]\(H_2\)[/tex].
- Calculate the moles of [tex]\(H_2\)[/tex] needed for the moles of [tex]\(N_2\)[/tex] available:
[tex]\[
\text{Moles of } H_2 \text{ needed} = 1.09 \, \text{moles of } N_2 \times 3 \approx 3.27 \, \text{moles}
\][/tex]
- Compare the available moles of [tex]\(H_2\)[/tex] (4.28 moles) to the moles needed (3.27 moles). Since 4.28 moles is greater than 3.27 moles, [tex]\(N_2\)[/tex] is the limiting reactant.
4. Calculate the Moles of [tex]\(NH_3\)[/tex] Formed:
- Use the limiting reactant [tex]\(N_2\)[/tex] to determine the [tex]\(NH_3\)[/tex] produced.
- According to the stoichiometry: [tex]\(1\)[/tex] mole of [tex]\(N_2\)[/tex] produces [tex]\(2\)[/tex] moles of [tex]\(NH_3\)[/tex].
[tex]\[
\text{Moles of } NH_3 = 1.09 \, \text{moles of } N_2 \times 2 = 2.18 \, \text{moles of } NH_3
\][/tex]
5. Calculate the Theoretical Yield of [tex]\(NH_3\)[/tex] in Grams:
- Finally, convert moles of [tex]\(NH_3\)[/tex] to grams:
[tex]\[
\text{Grams of } NH_3 = 2.18 \, \text{moles of } NH_3 \times 17.03 \, \text{g/mol} \approx 37.07 \, \text{g}
\][/tex]
Thus, the theoretical yield of [tex]\(NH_3\)[/tex] is approximately 37.1 g, matching the option "37.1 g NH_3".
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