Cuantos moles de NH3 se producen cuando un exceso de N2 reaccióna con 2?
Cuantos moles de NH3 se producen cuando un exceso de N2 reaccióna con 2. 00g de H2. N2 + H2 = > NH3.
Cuantos moles de NH3 se producen cuando un exceso de N2 reaccióna con 2. 00g de H2. N2 + H2 = > NH3.
Pregunta de Química · 1 respuesta · 5 votos · mejor respuesta de Sofivmgsofi2540
En resumen
N2 + 3 H2 → 2 NH3Mm NH3 = 17 g / mol N2 = 28 g / molcalcular moles de NH3n NH3 = 2. 00 g N2 x 1 mol N2 x 2 mol NH3 `````````````````` ``````````````````````` 28 g N2 1 mol N2n NH3 = 0. 142 moles.
N2 + 3 H2 → 2 NH3Mm NH3 = 17 g / mol N2 = 28 g / molcalcular moles de NH3n NH3 = 2.
00 g N2 x 1 mol N2 x 2 mol NH3 `````````````````` ``````````````````````` 28 g N2 1 mol N2n NH3 = 0.
142 moles.
N2 + 3 H2 → 2 NH3Mm NH3 = 17 g / mol N2 = 28 g / molcalcular moles de NH3n NH3 = 2. 00 g N2 x 1 mol N2 x 2 mol NH3 `````````````````` ``````````````````````` 28 g N2 1 mol N2n NH3 = 0. 142 moles.
2Al + 3Cl2 → 2AlCl3 Mm AlCl3 = 133 g / mol calcular moles de AlCl3 a partir de la reaccion : mol AlCl3 = 1. 5 mol Al x 2 mol AlCl3 ```````````````` 2 mol Al mol AlCl3 = 1. 5 mol AlCl3 mol AlCl3 = 3 mol Cl2 x 2 mol AlCl3…
2Al + 3Cl2 → 2AlCl3 a. Calcular moles a partir de la reacción : mol AlCl3 = 1. 5 mol Al x 2 mol AlCl3 `````````````````` 2 mol Al mol AlCl3 = 1. 5 mol AlCl3 = 3 mol Cl2 x 2 mol AlCl3 ```````````````` 3 mol Cl2 mol AlCl3…