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Answer :
To solve this problem, let's proceed step by step to find out how many grams of ammonia ([tex]$\text{NH}_3$[/tex]) are formed when 12.0 grams of hydrogen ([tex]$\text{H}_2$[/tex]) react with excess nitrogen ([tex]$\text{N}_2$[/tex]), given that the reaction yield is 85.0%.
1. Understand the Reaction:
The balanced chemical equation for the formation of ammonia is:
[tex]\[
\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)
\][/tex]
According to the equation, 3 moles of [tex]$\text{H}_2$[/tex] produce 2 moles of [tex]$\text{NH}_3$[/tex].
2. Calculate Moles of [tex]$\text{H}_2$[/tex]:
- The molar mass of [tex]$\text{H}_2$[/tex] is 2.0 g/mol.
- Compute the moles of [tex]$\text{H}_2$[/tex] from the given mass:
[tex]\[
\text{moles of } \text{H}_2 = \frac{12.0 \, \text{g}}{2.0 \, \text{g/mol}} = 6.0 \, \text{moles}
\][/tex]
3. Determine Moles of [tex]$\text{NH}_3$[/tex] Produced:
- From the stoichiometry of the reaction, 3 moles of [tex]$\text{H}_2$[/tex] produce 2 moles of [tex]$\text{NH}_3$[/tex].
- Therefore, the moles of [tex]$\text{NH}_3$[/tex] produced is:
[tex]\[
\text{moles of } \text{NH}_3 = \left(\frac{2}{3}\right) \times 6.0 \, \text{moles} = 4.0 \, \text{moles}
\][/tex]
4. Calculate Theoretical Mass of [tex]$\text{NH}_3$[/tex]:
- The molar mass of [tex]$\text{NH}_3$[/tex] is 17.0 g/mol.
- The theoretical mass of [tex]$\text{NH}_3$[/tex] is:
[tex]\[
\text{mass of } \text{NH}_3 = 4.0 \, \text{moles} \times 17.0 \, \text{g/mol} = 68.0 \, \text{g}
\][/tex]
5. Calculate Experimental Yield:
- Given that the yield of the reaction is 85.0%, the actual mass of [tex]$\text{NH}_3$[/tex] obtained is:
[tex]\[
\text{actual mass} = \left(\frac{85.0}{100}\right) \times 68.0 \, \text{g} = 57.8 \, \text{g}
\][/tex]
Therefore, the experimentally obtained mass of ammonia is 57.8 grams.
Matching this with the provided options, the correct answer is:
c. 57.2 g (Note: It appears there is a slight discrepancy here, but based on the calculation and closest option available, it should correspond to the experimental value given).
1. Understand the Reaction:
The balanced chemical equation for the formation of ammonia is:
[tex]\[
\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)
\][/tex]
According to the equation, 3 moles of [tex]$\text{H}_2$[/tex] produce 2 moles of [tex]$\text{NH}_3$[/tex].
2. Calculate Moles of [tex]$\text{H}_2$[/tex]:
- The molar mass of [tex]$\text{H}_2$[/tex] is 2.0 g/mol.
- Compute the moles of [tex]$\text{H}_2$[/tex] from the given mass:
[tex]\[
\text{moles of } \text{H}_2 = \frac{12.0 \, \text{g}}{2.0 \, \text{g/mol}} = 6.0 \, \text{moles}
\][/tex]
3. Determine Moles of [tex]$\text{NH}_3$[/tex] Produced:
- From the stoichiometry of the reaction, 3 moles of [tex]$\text{H}_2$[/tex] produce 2 moles of [tex]$\text{NH}_3$[/tex].
- Therefore, the moles of [tex]$\text{NH}_3$[/tex] produced is:
[tex]\[
\text{moles of } \text{NH}_3 = \left(\frac{2}{3}\right) \times 6.0 \, \text{moles} = 4.0 \, \text{moles}
\][/tex]
4. Calculate Theoretical Mass of [tex]$\text{NH}_3$[/tex]:
- The molar mass of [tex]$\text{NH}_3$[/tex] is 17.0 g/mol.
- The theoretical mass of [tex]$\text{NH}_3$[/tex] is:
[tex]\[
\text{mass of } \text{NH}_3 = 4.0 \, \text{moles} \times 17.0 \, \text{g/mol} = 68.0 \, \text{g}
\][/tex]
5. Calculate Experimental Yield:
- Given that the yield of the reaction is 85.0%, the actual mass of [tex]$\text{NH}_3$[/tex] obtained is:
[tex]\[
\text{actual mass} = \left(\frac{85.0}{100}\right) \times 68.0 \, \text{g} = 57.8 \, \text{g}
\][/tex]
Therefore, the experimentally obtained mass of ammonia is 57.8 grams.
Matching this with the provided options, the correct answer is:
c. 57.2 g (Note: It appears there is a slight discrepancy here, but based on the calculation and closest option available, it should correspond to the experimental value given).
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