An important process for the production of acrylonitrile (C3H3N) is given by the following equation:
2C3H6(g) + 2NH3(g) + 3O2(g) = 2C3H3N(g) + 6H2O(g)
A 150.-L reactor is charged to the following partial pressures at 21°C:
P of C3H6 = 0.540MPa
P of NH3 = 0.710 MPa
P of O2 = 1.580 MPa
What mass of acrylonitrile can be produced from this mixture (MPa = 10^6 Pa)?
Mass = _________g
chemistry problem help?
2017-10-18 8:07 am
回答 (2)
2017-10-18 8:40 am
2C₃H₆(g) + 2NH₃(g) + 3O₂(g) → 2C₃H₃N(g) + 6H₂O(g)
Mole ratio C₃H₆ : NH₃ : O₂ = 2 : 2 : 3
When 0.540 MPa C₃H₆ completely reacts :
NH₃ needed = (0.540 MPa) × (2/2) = 0.540 MPa < 0.710 MPa
O₂ needed = (0.540 MPa) × (3/2) = 0.810 MPa < 1.580 MPa
Hence, both NH₃ and O₂ are in excess, and C₃H₆ is the limiting reactant (completely reacts).
According to the above equation, mole ratio C₃H₆ : C₃H₃N = 2 : 2 = 1 : 1
Partial pressure of C₃H₆ reacted = 0.540 MPa
Partial pressure of C₃H₃N produced = 0.540 MPa
Consider the C₃H₃N gas produced :
Partial pressure, P = 0.540 MPa = 540 kPa
Volume, V = 150 L
Molar mass, M = (12.0×3 + 1.0×3 + 14.0) g/mol = 53.0 g/mol
Gas constant, R = 8.314 J / (mol K)
Temperature, T = (273 + 21) K = 294 K
PV = nRT and n = m/M
Then, PV = (m/M)RT
Hence, m = PVM/(RT)
Mass of C₃H₃N produced. m = 540 × 150 × 53.0 / (8.314 × 294) g = 1760 g
Mole ratio C₃H₆ : NH₃ : O₂ = 2 : 2 : 3
When 0.540 MPa C₃H₆ completely reacts :
NH₃ needed = (0.540 MPa) × (2/2) = 0.540 MPa < 0.710 MPa
O₂ needed = (0.540 MPa) × (3/2) = 0.810 MPa < 1.580 MPa
Hence, both NH₃ and O₂ are in excess, and C₃H₆ is the limiting reactant (completely reacts).
According to the above equation, mole ratio C₃H₆ : C₃H₃N = 2 : 2 = 1 : 1
Partial pressure of C₃H₆ reacted = 0.540 MPa
Partial pressure of C₃H₃N produced = 0.540 MPa
Consider the C₃H₃N gas produced :
Partial pressure, P = 0.540 MPa = 540 kPa
Volume, V = 150 L
Molar mass, M = (12.0×3 + 1.0×3 + 14.0) g/mol = 53.0 g/mol
Gas constant, R = 8.314 J / (mol K)
Temperature, T = (273 + 21) K = 294 K
PV = nRT and n = m/M
Then, PV = (m/M)RT
Hence, m = PVM/(RT)
Mass of C₃H₃N produced. m = 540 × 150 × 53.0 / (8.314 × 294) g = 1760 g
2017-10-18 8:41 am
2 C3H6 + 2 NH3 + 3 O2 → 2 C3H3N + 6 H2O
0.540 MPa of C3H6 would react completely with 0.540 x (2/2) = 0.540 MPa of NH3, but there is more NH3 present than that, so NH3 is in excess.
0.540 MPa of C3H6 would react completely with 0.540 x (3/2) = 0.810 MPa of O2, but there is more O2 present than that, so O2 is in excess.
Since NH3 and O2 are both in excess, that makes C3H6 the limiting reactant.
n = PV / RT = (540 kPa) x (150. L) / ((8.3144621 L kPa/K mol) x (21 + 273) K) = 33.136 mol C3H6
(33.136 mol C3H6) x (2 mol C3H3N / 2 mol C3H6) x (53.0626 C3H3N/mol) = 1758 g C3H3N
0.540 MPa of C3H6 would react completely with 0.540 x (2/2) = 0.540 MPa of NH3, but there is more NH3 present than that, so NH3 is in excess.
0.540 MPa of C3H6 would react completely with 0.540 x (3/2) = 0.810 MPa of O2, but there is more O2 present than that, so O2 is in excess.
Since NH3 and O2 are both in excess, that makes C3H6 the limiting reactant.
n = PV / RT = (540 kPa) x (150. L) / ((8.3144621 L kPa/K mol) x (21 + 273) K) = 33.136 mol C3H6
(33.136 mol C3H6) x (2 mol C3H3N / 2 mol C3H6) x (53.0626 C3H3N/mol) = 1758 g C3H3N
收錄日期: 2021-04-18 17:52:14
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