EJERCICIOS DE ELECTROQUIMICA
Enviado por Quijad1 • 5 de Junio de 2016 • Tarea • 1.708 Palabras (7 Páginas) • 158 Visitas
- H2O + C + H2SO4 ----------------→ H2SO3 + H2CO3 + H2O
3H2O + 2C + 4H2SO4 ----------------→ 4H2SO3 + 2H2CO3 + H2O
Obtener 85 gr de H2CO3
3H2O = 54 gr
2C = 24 gr
4H2SO4 = 392 gr
4H2SO3 = 328 gr
2H2CO3 = 124 gr
H2O = 18 gr
54 gr H2O ---------------------------------- 124 gr H2CO3
X ----------------------------------- 85 gr H2CO3
X = (85 gr H2CO3) (54 gr H2O)[pic 1]
(124 gr H2CO3) x = 37.0161 gr H2O
54 gr H2O ---------------------------------- 24 gr C
37.0161 gr H2O--------------------------- X
X = (37.0161 gr H2O) (24 gr C)[pic 2]
(54 gr H2O) x = 16.4516 gr C
54 gr H2O ---------------------------------- 392 gr H2SO4
37.0161 gr H2O--------------------------- X
X = (37.0161 gr H2O) (392 gr H2SO4)[pic 3]
(54 gr H2O) x = 268.7094 H2SO4
54 gr H2O ---------------------------------- 328 gr H2SO3
37.0161 gr H2O--------------------------- X
X = (37.0161 gr H2O) (328 gr H2SO3)[pic 4]
(54 gr H2O) x = 224.8385 gr H2SO3
54 gr H2O ---------------------------------- 18 gr H2O
37.0161 gr H2O--------------------------- X
X = (37.0161 gr H2O) (18 gr H2O)[pic 5]
(54 gr H2O) x = 12.3387 gr H2O
Comprobación
37.0161 gr 224.8385 gr
+ 16.4516 gr + 12.3387 gr
268.7094 gr 85.0000 gr[pic 6][pic 7]
322.1771 gr 322.1772 gr
- KMnO4 + HCl ----------------→ MnCl2 + Cl2 + KCl
2KMnO4 + 16HCl ----------------→ 2MnCl2 + 5Cl2 + 2KCl + 8H2O
Obtener 45 kg de KCl
2KMnO4 = 316.0726 kg
16HCl = 583.248 kg
2MnCl2 = 251.6882 kg
5Cl2 = 354.53 kg
2KCl = 149.11 kg
8H2O = 144 kg
316.0726 kg KMnO4 ---------------------------------- 149.11 kg KCl
X ----------------------------------- 45 kg KCl
X = (316.0726 kg KMnO4) (45 kg KCl)[pic 8]
(149.11 kg KCl) x = 95.3877 kg KMnO4
316.0726 kg KMnO4 --------------------------- 583.248 kg HCl
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