If 1.20 g of gaseous N2 react with 0.500 g of gaseous H2 in a 2.00-L closed container at 80.0 ℃, according to the equation shown below, what pressure does the gaseous NH3 exert against the container walls?
N2(g) + 3H2(g) → 2NH3(g)
Ap Chemistry Gas law help?
2020-10-05 10:08 pm
回答 (1)
2020-10-05 11:22 pm
Molar mass of N₂ = 14.0×2 g/mol = 28.0 g/mol
Molar mass of H₂ = 1.0×2 g/mol = 2.0 g/mol
Initial moles of N₂ = (1.20 g) / (28.0 g/mol) = 0.04286 mol
Initial moles of H₂ = (0.500 g) / (2.0 g/mol) = 0.25 mol
N₂ + 3H₂ → 2NH₃
Mole ratio N₂ : H₂ = 1 : 3
If N₂ completely reacts, H₂ needed = (0.04286 mol) × 3 = 0.1286 mol < 0.25 mol
Hence, H₂ is in excess, and N₂ is the limiting reactant.
According to the above equation, mole ratio N₂ : NH₃ = 1 : 2
Moles of NH₃ produced = (0.04286 mol) × 2 = 0.08572 mol
Consider the NH₃ gas produced:
Volume, V = 2.00 L
No. of moles, n = 0.08572 mol
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = (273.2 + 80.0) K = 353.2 K
Gas law: PV = nRT
Then, P = nRT/V
Pressure of NH₃, P = 0.08572 × 0.08206 × 353.2 / 2.00 atm = 1.24 atm
Molar mass of H₂ = 1.0×2 g/mol = 2.0 g/mol
Initial moles of N₂ = (1.20 g) / (28.0 g/mol) = 0.04286 mol
Initial moles of H₂ = (0.500 g) / (2.0 g/mol) = 0.25 mol
N₂ + 3H₂ → 2NH₃
Mole ratio N₂ : H₂ = 1 : 3
If N₂ completely reacts, H₂ needed = (0.04286 mol) × 3 = 0.1286 mol < 0.25 mol
Hence, H₂ is in excess, and N₂ is the limiting reactant.
According to the above equation, mole ratio N₂ : NH₃ = 1 : 2
Moles of NH₃ produced = (0.04286 mol) × 2 = 0.08572 mol
Consider the NH₃ gas produced:
Volume, V = 2.00 L
No. of moles, n = 0.08572 mol
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = (273.2 + 80.0) K = 353.2 K
Gas law: PV = nRT
Then, P = nRT/V
Pressure of NH₃, P = 0.08572 × 0.08206 × 353.2 / 2.00 atm = 1.24 atm
收錄日期: 2021-05-01 22:41:29
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