Respuesta :
Question:
a. 44 liters of NH3 are produced and 66 liters of H2 remain
b. 22 liters of NH3 are produced and no liters of H2 remain
c. 44 liters of NH3 are produced and no reactants remain
d. 22 liters of NH3 are produced and 66 liters of N2 remain
Answer:
The most correct option is;
a. 44 liters of NH₃ are produced and 66 liters of H₂ remain
Explanation:
We are given the equation of the reaction as follows;
3H₂ (g) + N₂ (g) → 2NH₃
Therefore, 3 moles of H₂ reacts with 1 mole of N₂ to produce 2 moles of NH₃
Molar mass of H₂ = 2.01588 g/mol
Molar mass of N₂ = 28.0134 g/mol
Molar mass of NH₃ = 17.031 g/mol
Therefore
Number of moles of H₂ = [tex]\frac{Mass \ of \ H_2}{Molar \ mass \ of \ H_2} = \frac{12}{2.01588 } = 5.953 \ moles[/tex]
∴ Number of moles of H₂ ≈ 6 moles
Number of moles of N₂ = [tex]\frac{Mass \ of \ N_2}{Molar \ mass \ of \ N_2} = \frac{28}{28.0134 } = 0.9995 \ moles[/tex]
∴ Number of moles of N₂ ≈ 1 mole
Since 3 moles of H₂ reacts with 1 mole of N₂ to produce 2 moles of NH₃, we have an excess of approximately 3 moles of H₂ with 2 moles of NH₃ produced
From the universal gas equation we have;
[tex]V = \frac{nRT}{P}[/tex]
Where:
For NH₃
n = Number of moles = 2
P = Pressure at STP = 1 atm
V = Volume = Required
T = Temperature at STP = 273.15 K
R = Universal Gas Constant =0.08205 L·atm/(mol·K)
Plugging the values, we have;
[tex]V = \frac{2 \times 0.08205 \times 273.15}{1} = 44.8 \, L[/tex]
For the excess H₂
n = Number of moles = 3
P = Pressure at STP = 1 atm
V = Volume = Required
T = Temperature at STP = 273.15 K
R = Universal Gas Constant =0.08205 L·atm/(mol·K)
Plugging the values, we have;
[tex]V = \frac{3 \times 0.08205 \times 273.15}{1} \approx 67 \, L[/tex]
Therefore, the most suitable option is 44 liters of NH₃ are produced and 66 liters of H₂ remain.
