Enunciado:
Se alimentan 50 mol de N₂ y 150 mol de H₂ para producir NH₃:
N₂ + 3 H₂ → 2 NH₃
Determinar el reactivo limitante, el exceso de H₂ y la composición si la conversión del N₂ es del 60%.
Solución paso a paso:
Review these topics so you can follow the solutions: Solucionario Ocon Tojo Capitulo 3
| Concept | Definition | Formula / Symbol | |---------|------------|------------------| | Stoichiometric coefficient | Number in front of species in balanced reaction | ( \nu_i ) | | Limiting reactant | The reactant that would run out first | ( \text(moles) / \nu_i ) smallest | | Excess reactant | Supplied more than stoichiometric amount | ( % \textexcess = \frac\textfeed - \textstoich\textstoich \times 100 ) | | Conversion (X) | Fraction of limiting reactant consumed | ( X = \frac\textreacted\textinitial ) | | Yield | Product obtained vs. theoretical max | ( \textYield = \frac\textactual product\textmax possible ) | | Selectivity | Desired product / undesired product | ( S = \frac\textmoles desired\textmoles undesired ) |
Problem: The reaction ( 2A + B \to C ) is fed with 100 mol A, 60 mol B. Conversion of A is 80%. Calculate all outlet moles. Enunciado : Se alimentan 50 mol de N₂
Solution (as solucionario would show):
Extent of reaction from A:
( X = 0.80 = \frac100 - n_A100 \Rightarrow n_A = 20 ) mol.
( n_A = n_A0 - 2\xi ) → ( 20 = 100 - 2\xi ) → ( \xi = 40 ) mol. Problem: The reaction ( 2A + B \to
Others:
Check total moles: 20+20+40 = 80 (initial 160, reaction reduces moles).
✅ Matches solucionario answer.